The article provides a detailed explanation of the definitions, the geometric meaning of the derivative with graphic notations. The equation of a tangent line will be considered with examples, the equations of a tangent to 2nd order curves will be found.
Definition 1The angle of inclination of the straight line y = k x + b is called angle α, which is measured from the positive direction of the x axis to the straight line y = k x + b in the positive direction.
In the figure, the x direction is indicated by a green arrow and a green arc, and the angle of inclination by a red arc. The blue line refers to the straight line.
Definition 2
The slope of the straight line y = k x + b is called the numerical coefficient k.
The angular coefficient is equal to the tangent of the straight line, in other words k = t g α.
- The angle of inclination of a straight line is equal to 0 only if it is parallel about x and the slope is equal to zero, because the tangent of zero is equal to 0. This means that the form of the equation will be y = b.
- If the angle of inclination of the straight line y = k x + b is acute, then the conditions 0 are satisfied< α < π 2 или 0 ° < α < 90 ° . Отсюда имеем, что значение углового коэффициента k считается положительным числом, потому как значение тангенс удовлетворяет условию t g α >0, and there is an increase in the graph.
- If α = π 2, then the location of the line is perpendicular to x. Equality is specified by x = c with the value c being a real number.
- If the angle of inclination of the straight line y = k x + b is obtuse, then it corresponds to the conditions π 2< α < π или 90 ° < α < 180 ° , значение углового коэффициента k принимает отрицательное значение, а график убывает.
A secant is a line that passes through 2 points of the function f (x). In other words, a secant is a straight line that is drawn through any two points on the graph of a given function.
The figure shows that A B is a secant, and f (x) is a black curve, α is a red arc, indicating the angle of inclination of the secant.
When the angular coefficient of a straight line is equal to the tangent of the angle of inclination, it is clear that the tangent of a right triangle A B C can be found by the ratio of the opposite side to the adjacent one.
Definition 4
We get a formula for finding a secant of the form:
k = t g α = B C A C = f (x B) - f x A x B - x A, where the abscissas of points A and B are the values x A, x B, and f (x A), f (x B) are the values functions at these points.
Obviously, the angular coefficient of the secant is determined using the equality k = f (x B) - f (x A) x B - x A or k = f (x A) - f (x B) x A - x B, and the equation must be written as y = f (x B) - f (x A) x B - x A x - x A + f (x A) or
y = f (x A) - f (x B) x A - x B x - x B + f (x B) .
The secant divides the graph visually into 3 parts: to the left of point A, from A to B, to the right of B. The figure below shows that there are three secants that are considered coincident, that is, they are set using a similar equation.
By definition, it is clear that the straight line and its secant in this case coincide.
A secant can intersect the graph of a given function multiple times. If there is an equation of the form y = 0 for a secant, then the number of points of intersection with the sinusoid is infinite.
Definition 5
Tangent to the graph of the function f (x) at point x 0 ; f (x 0) is a straight line passing through a given point x 0; f (x 0), with the presence of a segment that has many x values close to x 0.
Example 1
Let's take a closer look at the example below. Then it is clear that the line defined by the function y = x + 1 is considered tangent to y = 2 x at the point with coordinates (1; 2). For clarity, it is necessary to consider graphs with values close to (1; 2). The function y = 2 x is shown in black, the blue line is the tangent line, and the red dot is the intersection point.
Obviously, y = 2 x merges with the line y = x + 1.
To determine the tangent, we should consider the behavior of the tangent A B as point B approaches point A infinitely. For clarity, we present a drawing.
The secant A B, indicated by the blue line, tends to the position of the tangent itself, and the angle of inclination of the secant α will begin to tend to the angle of inclination of the tangent itself α x.
Definition 6
The tangent to the graph of the function y = f (x) at point A is considered to be the limiting position of the secant A B as B tends to A, that is, B → A.
Now let's move on to consider the geometric meaning of the derivative of a function at a point.
Let's move on to considering the secant A B for the function f (x), where A and B with coordinates x 0, f (x 0) and x 0 + ∆ x, f (x 0 + ∆ x), and ∆ x is denoted as the increment of the argument . Now the function will take the form ∆ y = ∆ f (x) = f (x 0 + ∆ x) - f (∆ x) . For clarity, let's give an example of a drawing.
Consider the resulting right triangle A B C. We use the definition of tangent to solve, that is, we obtain the relation ∆ y ∆ x = t g α . From the definition of a tangent it follows that lim ∆ x → 0 ∆ y ∆ x = t g α x . According to the rule of the derivative at a point, we have that the derivative f (x) at the point x 0 is called the limit of the ratio of the increment of the function to the increment of the argument, where ∆ x → 0, then we denote it as f (x 0) = lim ∆ x → 0 ∆ y ∆ x .
It follows that f " (x 0) = lim ∆ x → 0 ∆ y ∆ x = t g α x = k x, where k x is denoted as the slope of the tangent.
That is, we find that f ' (x) can exist at point x 0, and like the tangent to a given graph of the function at the point of tangency equal to x 0, f 0 (x 0), where the value of the slope of the tangent at the point is equal to the derivative at point x 0 . Then we get that k x = f " (x 0) .
The geometric meaning of the derivative of a function at a point is that it gives the concept of the existence of a tangent to the graph at the same point.
To write the equation of any straight line on a plane, it is necessary to have an angular coefficient with the point through which it passes. Its notation is taken to be x 0 at intersection.
The tangent equation to the graph of the function y = f (x) at the point x 0, f 0 (x 0) takes the form y = f "(x 0) x - x 0 + f (x 0).
This means that the final value of the derivative f "(x 0) can determine the position of the tangent, that is, vertically, provided lim x → x 0 + 0 f "(x) = ∞ and lim x → x 0 - 0 f "(x ) = ∞ or absence at all under the condition lim x → x 0 + 0 f " (x) ≠ lim x → x 0 - 0 f " (x) .
The location of the tangent depends on the value of its angular coefficient k x = f "(x 0). When parallel to the o x axis, we obtain that k k = 0, when parallel to o y - k x = ∞, and the form of the tangent equation x = x 0 increases with k x > 0, decreases as k x< 0 .
Example 2
Compile an equation for the tangent to the graph of the function y = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 at the point with coordinates (1; 3) and determine the angle of inclination.
Solution
By condition, we have that the function is defined for all real numbers. We find that the point with the coordinates specified by the condition, (1; 3) is a point of tangency, then x 0 = - 1, f (x 0) = - 3.
It is necessary to find the derivative at the point with value - 1. We get that
y " = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 " = = e x + 1 " + x 3 3 " - 6 - 3 3 x " - 17 - 3 3 " = e x + 1 + x 2 - 6 - 3 3 y " (x 0) = y " (- 1) = e - 1 + 1 + - 1 2 - 6 - 3 3 = 3 3
The value of f' (x) at the point of tangency is the slope of the tangent, which is equal to the tangent of the slope.
Then k x = t g α x = y " (x 0) = 3 3
It follows that α x = a r c t g 3 3 = π 6
Answer: the tangent equation takes the form
y = f " (x 0) x - x 0 + f (x 0) y = 3 3 (x + 1) - 3 y = 3 3 x - 9 - 3 3
For clarity, we give an example in a graphic illustration.
Black color is used for the graph of the original function, blue color is the image of the tangent, and the red dot is the point of tangency. The figure on the right shows an enlarged view.
Example 3
Determine the existence of a tangent to the graph of a given function
y = 3 · x - 1 5 + 1 at the point with coordinates (1 ; 1) . Write an equation and determine the angle of inclination.
Solution
By condition, we have that the domain of definition of a given function is considered to be the set of all real numbers.
Let's move on to finding the derivative
y " = 3 x - 1 5 + 1 " = 3 1 5 (x - 1) 1 5 - 1 = 3 5 1 (x - 1) 4 5
If x 0 = 1, then f' (x) is undefined, but the limits are written as lim x → 1 + 0 3 5 1 (x - 1) 4 5 = 3 5 1 (+ 0) 4 5 = 3 5 · 1 + 0 = + ∞ and lim x → 1 - 0 3 5 · 1 (x - 1) 4 5 = 3 5 · 1 (- 0) 4 5 = 3 5 · 1 + 0 = + ∞ , which means the existence vertical tangent at point (1; 1).
Answer: the equation will take the form x = 1, where the angle of inclination will be equal to π 2.
For clarity, let's depict it graphically.
Example 4
Find the points on the graph of the function y = 1 15 x + 2 3 - 4 5 x 2 - 16 5 x - 26 5 + 3 x + 2, where
- There is no tangent;
- The tangent is parallel to x;
- The tangent is parallel to the line y = 8 5 x + 4.
Solution
It is necessary to pay attention to the scope of definition. By condition, we have that the function is defined on the set of all real numbers. We expand the module and solve the system with intervals x ∈ - ∞ ; 2 and [ - 2 ; + ∞) . We get that
y = - 1 15 x 3 + 18 x 2 + 105 x + 176 , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 , x ∈ [ - 2 ; + ∞)
It is necessary to differentiate the function. We have that
y " = - 1 15 x 3 + 18 x 2 + 105 x + 176 " , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 ", x ∈ [ - 2 ; + ∞) ⇔ y " = - 1 5 (x 2 + 12 x + 35) , x ∈ - ∞ ; - 2 1 5 x 2 - 4 x + 3 , x ∈ [ - 2 ; + ∞)
When x = − 2, then the derivative does not exist because the one-sided limits are not equal at that point:
lim x → - 2 - 0 y " (x) = lim x → - 2 - 0 - 1 5 (x 2 + 12 x + 35 = - 1 5 (- 2) 2 + 12 (- 2) + 35 = - 3 lim x → - 2 + 0 y " (x) = lim x → - 2 + 0 1 5 (x 2 - 4 x + 3) = 1 5 - 2 2 - 4 - 2 + 3 = 3
We calculate the value of the function at the point x = - 2, where we get that
- y (- 2) = 1 15 - 2 + 2 3 - 4 5 (- 2) 2 - 16 5 (- 2) - 26 5 + 3 - 2 + 2 = - 2, that is, the tangent at the point (- 2; - 2) will not exist.
- The tangent is parallel to x when the slope is zero. Then k x = t g α x = f "(x 0). That is, it is necessary to find the values of such x when the derivative of the function turns it to zero. That is, the values of f ' (x) will be the points of tangency, where the tangent is parallel to x .
When x ∈ - ∞ ; - 2, then - 1 5 (x 2 + 12 x + 35) = 0, and for x ∈ (- 2; + ∞) we get 1 5 (x 2 - 4 x + 3) = 0.
1 5 (x 2 + 12 x + 35) = 0 D = 12 2 - 4 35 = 144 - 140 = 4 x 1 = - 12 + 4 2 = - 5 ∈ - ∞ ; - 2 x 2 = - 12 - 4 2 = - 7 ∈ - ∞ ; - 2 1 5 (x 2 - 4 x + 3) = 0 D = 4 2 - 4 · 3 = 4 x 3 = 4 - 4 2 = 1 ∈ - 2 ; + ∞ x 4 = 4 + 4 2 = 3 ∈ - 2 ; +∞
Calculate the corresponding function values
y 1 = y - 5 = 1 15 - 5 + 2 3 - 4 5 - 5 2 - 16 5 - 5 - 26 5 + 3 - 5 + 2 = 8 5 y 2 = y (- 7) = 1 15 - 7 + 2 3 - 4 5 (- 7) 2 - 16 5 - 7 - 26 5 + 3 - 7 + 2 = 4 3 y 3 = y (1) = 1 15 1 + 2 3 - 4 5 1 2 - 16 5 1 - 26 5 + 3 1 + 2 = 8 5 y 4 = y (3) = 1 15 3 + 2 3 - 4 5 3 2 - 16 5 3 - 26 5 + 3 3 + 2 = 4 3
Hence - 5; 8 5, - 4; 4 3, 1; 8 5, 3; 4 3 are considered to be the required points of the function graph.
Let's look at a graphical representation of the solution.
The black line is the graph of the function, the red dots are the tangency points.
- When the lines are parallel, the angular coefficients are equal. Then it is necessary to search for points on the function graph where the slope will be equal to the value 8 5. To do this, you need to solve an equation of the form y "(x) = 8 5. Then, if x ∈ - ∞; - 2, we obtain that - 1 5 (x 2 + 12 x + 35) = 8 5, and if x ∈ ( - 2 ; + ∞), then 1 5 (x 2 - 4 x + 3) = 8 5.
The first equation has no roots since the discriminant is less than zero. Let's write down that
1 5 x 2 + 12 x + 35 = 8 5 x 2 + 12 x + 43 = 0 D = 12 2 - 4 43 = - 28< 0
Another equation has two real roots, then
1 5 (x 2 - 4 x + 3) = 8 5 x 2 - 4 x - 5 = 0 D = 4 2 - 4 · (- 5) = 36 x 1 = 4 - 36 2 = - 1 ∈ - 2 ; + ∞ x 2 = 4 + 36 2 = 5 ∈ - 2 ; +∞
Let's move on to finding the values of the function. We get that
y 1 = y (- 1) = 1 15 - 1 + 2 3 - 4 5 (- 1) 2 - 16 5 (- 1) - 26 5 + 3 - 1 + 2 = 4 15 y 2 = y (5) = 1 15 5 + 2 3 - 4 5 5 2 - 16 5 5 - 26 5 + 3 5 + 2 = 8 3
Points with values - 1; 4 15, 5; 8 3 are the points at which the tangents are parallel to the line y = 8 5 x + 4.
Answer: black line – graph of the function, red line – graph of y = 8 5 x + 4, blue line – tangents at points - 1; 4 15, 5; 8 3.
There may be an infinite number of tangents for given functions.
Example 5
Write the equations of all available tangents of the function y = 3 cos 3 2 x - π 4 - 1 3, which are located perpendicular to the straight line y = - 2 x + 1 2.
Solution
To compile the tangent equation, it is necessary to find the coefficient and coordinates of the tangent point, based on the condition of perpendicularity of the lines. The definition is as follows: the product of angular coefficients that are perpendicular to straight lines is equal to - 1, that is, written as k x · k ⊥ = - 1. From the condition we have that the angular coefficient is located perpendicular to the line and is equal to k ⊥ = - 2, then k x = - 1 k ⊥ = - 1 - 2 = 1 2.
Now you need to find the coordinates of the touch points. You need to find x and then its value for a given function. Note that from the geometric meaning of the derivative at the point
x 0 we obtain that k x = y "(x 0). From this equality we find the values of x for the points of contact.
We get that
y " (x 0) = 3 cos 3 2 x 0 - π 4 - 1 3 " = 3 - sin 3 2 x 0 - π 4 3 2 x 0 - π 4 " = = - 3 sin 3 2 x 0 - π 4 3 2 = - 9 2 sin 3 2 x 0 - π 4 ⇒ k x = y " (x 0) ⇔ - 9 2 sin 3 2 x 0 - π 4 = 1 2 ⇒ sin 3 2 x 0 - π 4 = - 1 9
This trigonometric equation will be used to calculate the ordinates of the tangent points.
3 2 x 0 - π 4 = a r c sin - 1 9 + 2 πk or 3 2 x 0 - π 4 = π - a r c sin - 1 9 + 2 πk
3 2 x 0 - π 4 = - a r c sin 1 9 + 2 πk or 3 2 x 0 - π 4 = π + a r c sin 1 9 + 2 πk
x 0 = 2 3 π 4 - a r c sin 1 9 + 2 πk or x 0 = 2 3 5 π 4 + a r c sin 1 9 + 2 πk , k ∈ Z
Z is a set of integers.
x points of contact have been found. Now you need to move on to searching for the values of y:
y 0 = 3 cos 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - sin 2 3 2 x 0 - π 4 - 1 3 or y 0 = 3 - 1 - sin 2 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - - 1 9 2 - 1 3 or y 0 = 3 - 1 - - 1 9 2 - 1 3
y 0 = 4 5 - 1 3 or y 0 = - 4 5 + 1 3
From this we obtain that 2 3 π 4 - a r c sin 1 9 + 2 πk ; 4 5 - 1 3 , 2 3 5 π 4 + a r c sin 1 9 + 2 πk ; - 4 5 + 1 3 are the points of tangency.
Answer: the necessary equations will be written as
y = 1 2 x - 2 3 π 4 - a r c sin 1 9 + 2 πk + 4 5 - 1 3 , y = 1 2 x - 2 3 5 π 4 + a r c sin 1 9 + 2 πk - 4 5 + 1 3 , k ∈ Z
For a visual representation, consider a function and a tangent on a coordinate line.
The figure shows that the function is located on the interval [ - 10 ; 10 ], where the black line is the graph of the function, the blue lines are tangents, which are located perpendicular to the given line of the form y = - 2 x + 1 2. Red dots are touch points.
The canonical equations of 2nd order curves are not single-valued functions. Tangent equations for them are compiled according to known schemes.
Tangent to a circle
To define a circle with center at point x c e n t e r ; y c e n t e r and radius R, apply the formula x - x c e n t e r 2 + y - y c e n t e r 2 = R 2 .
This equality can be written as a union of two functions:
y = R 2 - x - x c e n t e r 2 + y c e n t e r y = - R 2 - x - x c e n t e r 2 + y c e n t e r
The first function is located at the top, and the second at the bottom, as shown in the figure.
To compile the equation of a circle at the point x 0; y 0 , which is located in the upper or lower semicircle, you should find the equation of the graph of a function of the form y = R 2 - x - x c e n t e r 2 + y c e n t e r or y = - R 2 - x - x c e n t e r 2 + y c e n t e r at the indicated point.
When at points x c e n t e r ; y c e n t e r + R and x c e n t e r ; y c e n t e r - R tangents can be given by the equations y = y c e n t e r + R and y = y c e n t e r - R , and at points x c e n t e r + R ; y c e n t e r and
x c e n t e r - R ; y c e n t e r will be parallel to o y, then we obtain equations of the form x = x c e n t e r + R and x = x c e n t e r - R .
Tangent to an ellipse
When the ellipse has a center at x c e n t e r ; y c e n t e r with semi-axes a and b, then it can be specified using the equation x - x c e n t e r 2 a 2 + y - y c e n t e r 2 b 2 = 1.
An ellipse and a circle can be denoted by combining two functions, namely the upper and lower half-ellipse. Then we get that
y = b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r y = - b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r
If the tangents are located at the vertices of the ellipse, then they are parallel about x or about y. Below, for clarity, consider the figure.
Example 6
Write the equation of the tangent to the ellipse x - 3 2 4 + y - 5 2 25 = 1 at points with values of x equal to x = 2.
Solution
It is necessary to find the tangent points that correspond to the value x = 2. We substitute into the existing equation of the ellipse and find that
x - 3 2 4 x = 2 + y - 5 2 25 = 1 1 4 + y - 5 2 25 = 1 ⇒ y - 5 2 = 3 4 25 ⇒ y = ± 5 3 2 + 5
Then 2 ; 5 3 2 + 5 and 2; - 5 3 2 + 5 are the tangent points that belong to the upper and lower half-ellipse.
Let's move on to finding and solving the equation of the ellipse with respect to y. We get that
x - 3 2 4 + y - 5 2 25 = 1 y - 5 2 25 = 1 - x - 3 2 4 (y - 5) 2 = 25 1 - x - 3 2 4 y - 5 = ± 5 1 - x - 3 2 4 y = 5 ± 5 2 4 - x - 3 2
Obviously, the upper half-ellipse is specified using a function of the form y = 5 + 5 2 4 - x - 3 2, and the lower half ellipse y = 5 - 5 2 4 - x - 3 2.
Let's apply a standard algorithm to create an equation for a tangent to the graph of a function at a point. Let us write that the equation for the first tangent at point 2; 5 3 2 + 5 will look like
y " = 5 + 5 2 4 - x - 3 2 " = 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = - 5 2 x - 3 4 - ( x - 3) 2 ⇒ y " (x 0) = y " (2) = - 5 2 2 - 3 4 - (2 - 3) 2 = 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = 5 2 3 (x - 2) + 5 3 2 + 5
We find that the equation of the second tangent with a value at the point
2 ; - 5 3 2 + 5 takes the form
y " = 5 - 5 2 4 - (x - 3) 2 " = - 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = 5 2 x - 3 4 - (x - 3) 2 ⇒ y " (x 0) = y " (2) = 5 2 2 - 3 4 - (2 - 3) 2 = - 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = - 5 2 3 (x - 2) - 5 3 2 + 5
Graphically, tangents are designated as follows:
Tangent to hyperbole
When a hyperbola has a center at x c e n t e r ; y c e n t e r and vertices x c e n t e r + α ; y c e n t e r and x c e n t e r - α ; y c e n t e r , the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = 1 takes place, if with vertices x c e n t e r ; y c e n t e r + b and x c e n t e r ; y c e n t e r - b , then is specified using the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = - 1 .
A hyperbola can be represented as two combined functions of the form
y = b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r or y = b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r
In the first case we have that the tangents are parallel to y, and in the second they are parallel to x.
It follows that in order to find the equation of the tangent to a hyperbola, it is necessary to find out which function the point of tangency belongs to. To determine this, it is necessary to substitute into the equations and check for identity.
Example 7
Write an equation for the tangent to the hyperbola x - 3 2 4 - y + 3 2 9 = 1 at point 7; - 3 3 - 3 .
Solution
It is necessary to transform the solution record for finding a hyperbola using 2 functions. We get that
x - 3 2 4 - y + 3 2 9 = 1 ⇒ y + 3 2 9 = x - 3 2 4 - 1 ⇒ y + 3 2 = 9 x - 3 2 4 - 1 ⇒ y + 3 = 3 2 x - 3 2 - 4 and y + 3 = - 3 2 x - 3 2 - 4 ⇒ y = 3 2 x - 3 2 - 4 - 3 y = - 3 2 x - 3 2 - 4 - 3
It is necessary to identify which function a given point with coordinates 7 belongs to; - 3 3 - 3 .
Obviously, to check the first function it is necessary y (7) = 3 2 · (7 - 3) 2 - 4 - 3 = 3 3 - 3 ≠ - 3 3 - 3, then the point does not belong to the graph, since the equality does not hold.
For the second function we have that y (7) = - 3 2 · (7 - 3) 2 - 4 - 3 = - 3 3 - 3 ≠ - 3 3 - 3, which means the point belongs to the given graph. From here you should find the slope.
We get that
y " = - 3 2 (x - 3) 2 - 4 - 3 " = - 3 2 x - 3 (x - 3) 2 - 4 ⇒ k x = y " (x 0) = - 3 2 x 0 - 3 x 0 - 3 2 - 4 x 0 = 7 = - 3 2 7 - 3 7 - 3 2 - 4 = - 3
Answer: the tangent equation can be represented as
y = - 3 x - 7 - 3 3 - 3 = - 3 x + 4 3 - 3
It is clearly depicted like this:
Tangent to a parabola
To create an equation for the tangent to the parabola y = a x 2 + b x + c at the point x 0, y (x 0), you must use a standard algorithm, then the equation will take the form y = y "(x 0) x - x 0 + y ( x 0).Such a tangent at the vertex is parallel to x.
You should define the parabola x = a y 2 + b y + c as the union of two functions. Therefore, we need to solve the equation for y. We get that
x = a y 2 + b y + c ⇔ a y 2 + b y + c - x = 0 D = b 2 - 4 a (c - x) y = - b + b 2 - 4 a (c - x) 2 a y = - b - b 2 - 4 a (c - x) 2 a
Graphically depicted as:
To find out whether a point x 0, y (x 0) belongs to a function, proceed gently according to the standard algorithm. Such a tangent will be parallel to o y relative to the parabola.
Example 8
Write the equation of the tangent to the graph x - 2 y 2 - 5 y + 3 when we have a tangent angle of 150 °.
Solution
We begin the solution by representing the parabola as two functions. We get that
2 y 2 - 5 y + 3 - x = 0 D = (- 5) 2 - 4 · (- 2) · (3 - x) = 49 - 8 x y = 5 + 49 - 8 x - 4 y = 5 - 49 - 8 x - 4
The value of the slope is equal to the value of the derivative at point x 0 of this function and is equal to the tangent of the angle of inclination.
We get:
k x = y "(x 0) = t g α x = t g 150 ° = - 1 3
From here we determine the x value for the points of contact.
The first function will be written as
y " = 5 + 49 - 8 x - 4 " = 1 49 - 8 x ⇒ y " (x 0) = 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3
Obviously, there are no real roots, since we got a negative value. We conclude that there is no tangent with an angle of 150° for such a function.
The second function will be written as
y " = 5 - 49 - 8 x - 4 " = - 1 49 - 8 x ⇒ y " (x 0) = - 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3 x 0 = 23 4 ⇒ y (x 0) = 5 - 49 - 8 23 4 - 4 = - 5 + 3 4
We have that the points of contact are 23 4 ; - 5 + 3 4 .
Answer: the tangent equation takes the form
y = - 1 3 x - 23 4 + - 5 + 3 4
Let's depict it graphically this way:
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A straight line that has only one common point with a circle is called a tangent to the circle, and their common point is called the tangent point of the line and the circle.
Theorem (property of a tangent to a circle)
A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
Given
A – point of contact
Prove:p OA
Proof.
Let's prove it by contradiction.
Let us assume that p is OA, then OA is inclined to the straight line p.
If from point O we draw a perpendicular OH to straight line p, then its length will be less than the radius: OH< ОА=r
We find that the distance from the center of the circle to the straight line p (OH) is less than the radius (r), which means the straight line p is secant (that is, it has two common points with the circle), which contradicts the conditions of the theorem (p is tangent).
This means that the assumption is incorrect, therefore the straight line p is perpendicular to OA.
Theorem (Property of tangent segments drawn from one point)
Segments of tangents to a circle drawn from one point are equal and make equal angles with a straight line passing through this point and the center of the circle. 
Given: approx. (O;r)
AB and AC are tangents to the surroundings. (O;r)
Prove: AB=AC
Proof
1) OB AB, OS AC, as radii drawn to the point of tangency (tangent property)
2) Consider tr. AOB, etc. AOS – p/u
JSC – general
OB=OS (as radii)
This means ABO = AOC (by hypotenuse and leg). Hence,
AB = AC,<3 = < 4 (как соответственные элементы в равных тр-ках). ч.т.д.
Theorem (Tangential test)
If a line passes through the end of a radius lying on a circle and is perpendicular to this radius, then it is a tangent. 
Given: OA – radius of the circle
Prove: p- tangent to the circle
Proof
OA – radius of the circle (according to condition) (OA=r)
OA – perpendicular from O to straight line p (OA =d)
This means that r=OA=d, which means that the straight line p and the circle have one common point.
Therefore, line p is tangent to the circle. etc.
3.Properties of chords and secants.
Properties of tangent and secant
DEFINITION
Circumference is the locus of points equidistant from one point, which is called the center of the circle.
A line segment connecting two points on a circle is called chord(in the figure this is a segment). A chord passing through the center of a circle is called diameter circles.
1. The tangent is perpendicular to the radius drawn to the point of contact.
2. Tangent segments drawn from one point are equal.
3. If a tangent and a secant are drawn from a point lying outside the circle, then the square of the length of the tangent is equal to the product of the secant and its outer part.
Lesson Objectives
- Educational – repetition, generalization and testing of knowledge on the topic: “Tangent to a circle”; development of basic skills.
- Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
- Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.
- Introduce the concept of a tangent, a point of contact.
- Consider the property of a tangent and its sign and show their application in solving problems in nature and technology.
Lesson Objectives
- Develop skills in constructing tangents using a scale ruler, protractor and drawing triangle.
- Test students' problem-solving skills.
- Ensure mastery of the basic algorithmic techniques for constructing a tangent to a circle.
- Develop the ability to apply theoretical knowledge to problem solving.
- Develop students' thinking and speech.
- Work on developing the skills to observe, notice patterns, generalize, and reason by analogy.
- Instilling an interest in mathematics.
Lesson Plan
- The emergence of the concept of tangent.
- The history of the tangent.
- Geometric definitions.
- Basic theorems.
- Constructing a tangent to a circle.
- Consolidation.
The emergence of the concept of tangent
The concept of a tangent is one of the oldest in mathematics. In geometry, a tangent to a circle is defined as a line that has exactly one point of intersection with that circle. The ancients, using compasses and rulers, were able to draw tangents to a circle, and later to conic sections: ellipses, hyperbolas and parabolas.
The history of the tangent
Interest in tangents was revived in modern times. Then curves were discovered that were unknown to ancient scientists. For example, Galileo introduced the cycloid, and Descartes and Fermat constructed a tangent to it. In the first third of the 17th century. They began to understand that a tangent is a straight line, “most closely adjacent” to a curve in a small neighborhood of a given point. It is easy to imagine a situation where it is impossible to construct a tangent to the curve at a given point (figure). 
Geometric definitions
Circle- the geometric locus of points on the plane equidistant from a given point, called its center.
circle.
Related definitions
- A segment connecting the center of a circle with any point on it (as well as the length of this segment) is called radius circles.
- The part of the plane bounded by a circle is called all around.
- A segment connecting two points on a circle is called its chord. A chord passing through the center of a circle is called diameter.
- Any two divergent points on a circle divide it into two parts. Each of these parts is called arc circles. The measure of an arc can be the measure of its corresponding central angle. An arc is called a semicircle if the segment connecting its ends is a diameter.
- A straight line that has exactly one common point with a circle is called tangent to a circle, and their common point is called the tangency point of the line and the circle.
- A straight line passing through two points on a circle is called secant.
- A central angle in a circle is a plane angle with a vertex at its center.
- An angle whose vertex lies on a circle and whose sides intersect this circle is called inscribed angle.
- Two circles having a common center are called concentric.
Tangent line- a straight line passing through a point on a curve and coinciding with it at this point up to first order.
Tangent to a circle is a straight line that has one common point with a circle.
A straight line passing through a point on a circle in the same plane perpendicular to the radius drawn to this point called tangent. In this case, this point on the circle is called the point of tangency.
Where in our cases “a” is a straight line which is tangent to a given circle, point “A” is the point of tangency. In this case, a⊥OA (straight line a is perpendicular to the radius OA).
They say that two circles touch, if they have a single common point. This point is called point of contact of the circles. Through the point of contact, you can draw a tangent to one of the circles, which is also a tangent to the other circle. Touching circles can be internal or external.
A tangency is called internal if the centers of the circles lie on the same side of the tangent.
A tangency is called external if the centers of the circles lie on opposite sides of the tangent
a is the common tangent to the two circles, K is the point of tangency.
Basic theorems
Theorem about tangent and secant
If a tangent and a secant are drawn from a point lying outside the circle, then the square of the length of the tangent is equal to the product of the secant and its outer part: MC 2 = MA MB.
Theorem. The radius drawn to the point of tangency of the circle is perpendicular to the tangent.
Theorem. If the radius is perpendicular to a line at the point where it intersects a circle, then this line is tangent to this circle.
Proof.
To prove these theorems, we need to remember what a perpendicular from a point to a line is. This is the shortest distance from this point to this line. Let us assume that OA is not perpendicular to the tangent, but there is a straight line OS perpendicular to the tangent. The length OS includes the length of the radius and a certain segment BC, which is certainly greater than the radius. Thus, one can prove it for any line. We conclude that the radius, the radius drawn to the point of contact, is the shortest distance to the tangent from point O, i.e. OS is perpendicular to the tangent. In the proof of the converse theorem, we will proceed from the fact that the tangent has only one common point with the circle. Let this straight line have one more common point B with the circle. Triangle AOB is rectangular and its two sides are equal as radii, which cannot be the case. Thus, we find that this straight line has no more points in common with the circle except point A, i.e. is tangent.
Theorem. The tangent segments drawn from one point to the circle are equal, and the straight line connecting this point with the center of the circle divides the angle between the tangents.
Proof.
The proof is very simple. Using the previous theorem, we assert that OB is perpendicular to AB, and OS is perpendicular to AC. Right triangles ABO and ACO are equal in leg and hypotenuse (OB=OS - radii, AO - total). Therefore, their sides AB=AC and angles OAC and OAB are equal.
Theorem. The magnitude of the angle formed by a tangent and a chord having a common point on a circle is equal to half the angular magnitude of the arc enclosed between its sides.
Proof.
Consider the angle NAB formed by a tangent and a chord. Let's draw the diameter of AC. The tangent is perpendicular to the diameter drawn to the point of contact, therefore, ∠CAN=90 o. Knowing the theorem, we see that angle alpha (a) is equal to half the angular value of the arc BC or half the angle BOS. ∠NAB=90 o -a, from here we get ∠NAB=1/2(180 o -∠BOC)=1/2∠AOB or = half the angular value of the arc BA. etc.
Theorem. If a tangent and a secant are drawn from a point to a circle, then the square of the tangent segment from a given point to the point of tangency is equal to the product of the lengths of the secant segments from a given point to the points of its intersection with the circle.
Proof.
In the figure, this theorem looks like this: MA 2 = MV * MC. Let's prove it. According to the previous theorem, the angle MAC is equal to half the angular value of the arc AC, but also the angle ABC is equal to half the angular value of the arc AC according to the theorem, therefore, these angles are equal to each other. Taking into account the fact that triangles AMC and BMA have a common angle at the vertex M, we state the similarity of these triangles in two angles (second sign). From the similarity we have: MA/MB=MC/MA, from which we get MA 2 =MB*MC
Constructing tangents to a circle
Now let's try to figure it out and find out what needs to be done to construct a tangent to a circle.
In this case, as a rule, the problem gives a circle and a point. And you and I need to construct a tangent to the circle so that this tangent passes through a given point.
In the event that we do not know the location of a point, then let's consider cases of possible locations of points.
Firstly, a point may be inside a circle, which is limited by a given circle. In this case, it is not possible to construct a tangent through this circle.
In the second case, the point is located on a circle, and we can construct a tangent by drawing a perpendicular line to the radius, which is drawn to the point known to us.
Thirdly, let’s assume that the point is located outside the circle, which is limited by the circle. In this case, before constructing a tangent, it is necessary to find a point on the circle through which the tangent must pass.
With the first case, I hope everything is clear to you, but to solve the second option we need to construct a segment on the straight line on which the radius lies. This segment must be equal to the radius and the segment that lies on the circle on the opposite side.
Here we see that a point on a circle is the middle of a segment that is equal to twice the radius. The next step will be to construct two circles. The radii of these circles will be equal to twice the radius of the original circle, with centers at the ends of the segment, which is equal to twice the radius. Now we can draw a straight line through any point of intersection of these circles and a given point. Such a straight line is the median perpendicular to the radius of the circle that was drawn initially. Thus, we see that this line is perpendicular to the circle and it follows from this that it is tangent to the circle.
In the third option, we have a point lying outside the circle, which is limited by a circle. In this case, we first construct a segment that will connect the center of the provided circle and the given point. And then we find its middle. But for this it is necessary to construct a perpendicular bisector. And you already know how to build it. Then we need to draw a circle, or at least part of it. Now we see that the point of intersection of the given circle and the newly constructed one is the point through which the tangent passes. It also passes through the point that was specified according to the conditions of the problem. And finally, through the two points you know, you can draw a tangent line.
And finally, in order to prove that the straight line we constructed is a tangent, we need to pay attention to the angle that was formed by the radius of the circle and the segment known by the condition and connecting the point of intersection of the circles with the point given by the condition of the problem. Now we see that the resulting angle rests on a semicircle. And from this it follows that this angle is right. Consequently, the radius will be perpendicular to the newly constructed line, and this line is the tangent.
Construction of a tangent.
The construction of tangent lines is one of those problems that led to the birth of differential calculus. The first published work related to differential calculus, written by Leibniz, was entitled “A new method of maxima and minima, as well as tangents, for which neither fractional nor irrational quantities, nor a special type of calculus, are an obstacle.”
Geometric knowledge of the ancient Egyptians.
If we do not take into account the very modest contribution of the ancient inhabitants of the valley between the Tigris and Euphrates and Asia Minor, then geometry originated in Ancient Egypt before 1700 BC. During the tropical rainy season, the Nile replenished its water reserves and overflowed. Water covered areas of cultivated land, and for tax purposes it was necessary to determine how much land was lost. Surveyors used a tightly stretched rope as a measuring tool. Another incentive for the accumulation of geometric knowledge by the Egyptians was their activities such as the construction of pyramids and fine arts.
The level of geometric knowledge can be judged from ancient manuscripts, which are specifically devoted to mathematics and are something like textbooks, or rather, problem books, where solutions to various practical problems are given.
The oldest mathematical manuscript of the Egyptians was copied by a certain student between 1800 - 1600. BC. from an older text. The papyrus was found by the Russian Egyptologist Vladimir Semenovich Golenishchev. It is kept in Moscow - in the Museum of Fine Arts named after A.S. Pushkin, and is called the Moscow papyrus.
Another mathematical papyrus, written two to three hundred years later than Moscow’s, is kept in London. It is called: “Instruction on how to achieve knowledge of all dark things, all the secrets that things hide in themselves... According to old monuments, the scribe Ahmes wrote this.” The manuscript is called the “Ahmes papyrus”, or the Rhind papyrus - after the name of the Englishman who found and bought this papyrus in Egypt. The Ahmes papyrus provides solutions to 84 problems involving various calculations that may be needed in practice.
A straight line relative to a circle can be in the following three positions:- The distance from the center of the circle to the straight line is greater than the radius. In this case, all points of the line lie outside the circle.
- The distance from the center of the circle to the straight line is less than the radius. In this case, the straight line has points lying inside the circle and since the straight line is infinite in both directions, it is intersected by the circle at 2 points.
- The distance from the center of the circle to the straight line is equal to the radius. Straight line is tangent.
A straight line that has only one point in common with a circle is called tangent to the circle.
The common point is called in this case point of contact.
The possibility of the existence of a tangent, and, moreover, drawn through any point of the circle as a point of tangency, is proved by the following theorem.
Theorem. If a line is perpendicular to the radius at its end lying on the circle, then this line is a tangent.
Let O (fig) be the center of some circle and OA some of its radius. Through its end A we draw MN ^ OA.
It is required to prove that the line MN is tangent, i.e. that this line has only one common point A with the circle.
Let us assume the opposite: let MN have another common point with the circle, for example B.
Then straight line OB would be a radius and therefore equal to OA.
But this cannot be, since if OA is perpendicular, then OB must be inclined to MN, and the inclined one is greater than the perpendicular.
Converse theorem. If a line is tangent to a circle, then the radius drawn to the point of tangency is perpendicular to it.
Let MN be the tangent to the circle, A the point of tangency, and O the center of the circle.
It is required to prove that OA^MN.
Let's assume the opposite, i.e. Let us assume that the perpendicular dropped from O to MN will not be OA, but some other line, for example, OB.
Let's take BC = AB and carry out OS.
Then OA and OS will be inclined, equally distant from the perpendicular OB, and therefore OS = OA.
It follows from this that the circle, taking into account our assumption, will have two common points with the line MN: A and C, i.e. MN will not be a tangent, but a secant, which contradicts the condition.
Consequence. Through any given point on a circle one can draw a tangent to this circle, and only one, since through this point one can draw a perpendicular, and only one, to the radius drawn into it.
Theorem. A tangent parallel to a chord divides the arc subtended by the chord in half at the point of contact.
Let straight line AB (fig.) touch the circle at point M and be parallel to chord CD.
We need to prove that ÈCM = ÈMD.
Drawing the diameter ME through the point of tangency, we obtain: EM ^ AB, and therefore EM ^ CB.
Therefore CM=MD.
Task. Through a given point draw a tangent to a given circle.
If a given point is on a circle, then draw a radius through it and a perpendicular straight line through the end of the radius. This line will be the desired tangent.
Let us consider the case when the point is given outside the circle.
Let it be required (Fig.) to draw a tangent to a circle with center O through point A.
To do this, from point A, as the center, we describe an arc with radius AO, and from point O, as the center, we intersect this arc at points B and C with a compass opening equal to the diameter of the given circle.
Having then drawn the chords OB and OS, we connect point A with points D and E, at which these chords intersect with the given circle.
Lines AD and AE are tangent to circle O.
Indeed, from the construction it is clear that the pipes AOB and AOC are isosceles (AO = AB = AC) with the bases OB and OS equal to the diameter of the circle O.
Since OD and OE are radii, then D is the middle of OB, and E is the middle of OS, which means AD and AE are medians drawn to the bases of isosceles pipes, and therefore perpendicular to these bases. If the lines DA and EA are perpendicular to the radii OD and OE, then they are tangent.
Consequence. Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.
So AD=AE and ÐOAD = ÐOAE (Fig.), because rectangular tr-ki AOD and AOE, having a common hypotenuse AO and equal legs OD and OE (as radii), are equal.
Note that here the word “tangent” means the actual “tangent segment” from a given point to the point of contact.
Task. Draw a tangent to a given circle O parallel to a given straight line AB (Fig.).
We lower a perpendicular OS to AB from the center O and through the point D, at which this perpendicular intersects the circle, draw EF || AB.
The tangent we are looking for will be EF.
Indeed, since OS ^ AB and EF || AB, then EF ^ OD, and the line perpendicular to the radius at its end lying on the circle is a tangent.
Task. Draw a common tangent to two circles O and O 1 (Fig.).
Analysis. Let's assume that the problem is solved.
Let AB be the common tangent, A and B the points of tangency.
Obviously, if we find one of these points, for example, A, then we can easily find the other one.
Let us draw the radii OA and O 1 B. These radii, being perpendicular to the common tangent, are parallel to each other.
Therefore, if from O 1 we draw O 1 C || BA, then the pipeline OCO 1 will be rectangular at vertex C.
As a result, if we describe a circle from O as the center with radius OS, then it will touch the straight line O 1 C at point C.
The radius of this auxiliary circle is known: it is equal to OA – CA = OA - O 1 B, i.e. it is equal to the difference between the radii of these circles.
Construction. From the center O we describe a circle with a radius equal to the difference of these radii.
From O 1 we draw a tangent O 1 C to this circle (in the manner indicated in the previous problem).
Through the tangent point C we draw the radius OS and continue it until it meets the given circle at point A. Finally, from A we draw AB parallel to CO 1.
In exactly the same way we can construct another common tangent A 1 B 1 (Fig.). Direct lines AB and A 1 B 1 are called external common tangents.
You can spend two more internal tangents as follows:
Analysis. Let's assume that the problem is solved (Fig.). Let AB be the desired tangent.
Let us draw the radii OA and O 1 B to the tangent points A and B. Since these radii are both perpendicular to the common tangent, they are parallel to each other.
Therefore, if from O 1 we draw O 1 C || BA and continue OA to point C, then OS will be perpendicular to O 1 C.
As a result, the circle described by the radius OS from point O as the center will touch the straight line O 1 C at point C.
The radius of this auxiliary circle is known: it is equal to OA+AC = OA+O 1 B, i.e. it is equal to the sum of the radii of the given circles.
Construction. From O as the center, we describe a circle with a radius equal to the sum of these radii.
From O 1 we draw a tangent O 1 C to this circle.
We connect the point of contact C with O.
Finally, through point A, at which OS intersects the given circle, we draw AB = O 1 C.
In a similar way we can construct another internal tangent A 1 B 1.
General definition of tangent
Let a tangent AT and some secant AM be drawn through point A to a circle with a center (Fig.).
Let's rotate this secant around point A so that the other intersection point B moves closer and closer to A.
Then the perpendicular OD, lowered from the center to the secant, will approach the radius OA more and more, and the angle AOD may become less than any small angle.
The angle MAT formed by the secant and tangent is equal to the angle AOD (due to the perpendicularity of their sides).
Therefore, as point B approaches A indefinitely, angle MAT can also become arbitrarily small.
This is expressed in other words like this:
a tangent is the limiting position to which a secant drawn through a point of tangency tends when the second point of intersection approaches the point of tangency indefinitely.
This property is taken as the definition of a tangent when talking about any curve.
Thus, the tangent to the curve AB (Fig.) is the limiting position MT to which the secant MN tends when the intersection point P approaches M without limit.
Note that the tangent defined in this way can have more than one common point with the curve (as can be seen in Fig.).
\[(\Large(\text(Central and inscribed angles)))\]
Definitions
A central angle is an angle whose vertex lies at the center of the circle.
An inscribed angle is an angle whose vertex lies on a circle.
The degree measure of an arc of a circle is the degree measure of the central angle that subtends it.
Theorem
The degree measure of an inscribed angle is equal to half the degree measure of the arc on which it rests.
Proof
We will carry out the proof in two stages: first, we will prove the validity of the statement for the case when one of the sides of the inscribed angle contains a diameter. Let point \(B\) be the vertex of the inscribed angle \(ABC\) and \(BC\) be the diameter of the circle:
Triangle \(AOB\) is isosceles, \(AO = OB\) , \(\angle AOC\) is external, then \(\angle AOC = \angle OAB + \angle ABO = 2\angle ABC\), where \(\angle ABC = 0.5\cdot\angle AOC = 0.5\cdot\buildrel\smile\over(AC)\).
Now consider an arbitrary inscribed angle \(ABC\) . Let us draw the diameter of the circle \(BD\) from the vertex of the inscribed angle. There are two possible cases:
1) the diameter cuts the angle into two angles \(\angle ABD, \angle CBD\) (for each of which the theorem is true as proven above, therefore it is also true for the original angle, which is the sum of these two and therefore equal to half the sum of the arcs to which they rest, that is, equal to half the arc on which it rests). Rice. 1.
2) the diameter did not cut the angle into two angles, then we have two more new inscribed angles \(\angle ABD, \angle CBD\), whose side contains the diameter, therefore, the theorem is true for them, then it is also true for the original angle (which is equal to the difference of these two angles, which means it is equal to the half-difference of the arcs on which they rest, that is, equal to half the arc on which it rests). Rice. 2.
Consequences
1. Inscribed angles subtending the same arc are equal.
2. An inscribed angle subtended by a semicircle is a right angle.
3. An inscribed angle is equal to half the central angle subtended by the same arc.
\[(\Large(\text(Tangent to the circle)))\]
Definitions
There are three types of relative positions of a line and a circle:
1) straight line \(a\) intersects the circle at two points. Such a line is called a secant line. In this case, the distance \(d\) from the center of the circle to the straight line is less than the radius \(R\) of the circle (Fig. 3).
2) straight line \(b\) intersects the circle at one point. Such a line is called a tangent, and their common point \(B\) is called the point of tangency. In this case \(d=R\) (Fig. 4).
Theorem
1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
2. If a line passes through the end of the radius of a circle and is perpendicular to this radius, then it is tangent to the circle.
Consequence
The tangent segments drawn from one point to a circle are equal.
Proof
Let us draw two tangents \(KA\) and \(KB\) to the circle from the point \(K\):
This means that \(OA\perp KA, OB\perp KB\) are like radii. Right triangles \(\triangle KAO\) and \(\triangle KBO\) are equal in leg and hypotenuse, therefore, \(KA=KB\) .
Consequence
The center of the circle \(O\) lies on the bisector of the angle \(AKB\) formed by two tangents drawn from the same point \(K\) .
\[(\Large(\text(Theorems related to angles)))\]
Theorem on the angle between secants
The angle between two secants drawn from the same point is equal to the half-difference in degree measures of the larger and smaller arcs they cut.
Proof
Let \(M\) be the point from which two secants are drawn as shown in the figure:
Let's show that \(\angle DMB = \dfrac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\).
\(\angle DAB\) is the external angle of the triangle \(MAD\), then \(\angle DAB = \angle DMB + \angle MDA\), where \(\angle DMB = \angle DAB - \angle MDA\), but the angles \(\angle DAB\) and \(\angle MDA\) are inscribed, then \(\angle DMB = \angle DAB - \angle MDA = \frac(1)(2)\buildrel\smile\over(BD) - \frac(1)(2)\buildrel\smile\over(CA) = \frac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\), which was what needed to be proven.
Theorem on the angle between intersecting chords
The angle between two intersecting chords is equal to half the sum of the degree measures of the arcs they cut: \[\angle CMD=\dfrac12\left(\buildrel\smile\over(AB)+\buildrel\smile\over(CD)\right)\]
Proof
\(\angle BMA = \angle CMD\) as vertical.
From triangle \(AMD\) : \(\angle AMD = 180^\circ - \angle BDA - \angle CAD = 180^\circ - \frac12\buildrel\smile\over(AB) - \frac12\buildrel\smile\over(CD)\).
But \(\angle AMD = 180^\circ - \angle CMD\), from which we conclude that \[\angle CMD = \frac12\cdot\buildrel\smile\over(AB) + \frac12\cdot\buildrel\smile\over(CD) = \frac12(\buildrel\smile\over(AB) + \buildrel\ smile\over(CD)).\]
Theorem on the angle between a chord and a tangent
The angle between the tangent and the chord passing through the point of tangency is equal to half the degree measure of the arc subtended by the chord.
Proof
Let the straight line \(a\) touch the circle at the point \(A\), \(AB\) is the chord of this circle, \(O\) is its center. Let the line containing \(OB\) intersect \(a\) at the point \(M\) . Let's prove that \(\angle BAM = \frac12\cdot \buildrel\smile\over(AB)\).
Let's denote \(\angle OAB = \alpha\) . Since \(OA\) and \(OB\) are radii, then \(OA = OB\) and \(\angle OBA = \angle OAB = \alpha\). Thus, \(\buildrel\smile\over(AB) = \angle AOB = 180^\circ - 2\alpha = 2(90^\circ - \alpha)\).
Since \(OA\) is the radius drawn to the tangent point, then \(OA\perp a\), that is, \(\angle OAM = 90^\circ\), therefore, \(\angle BAM = 90^\circ - \angle OAB = 90^\circ - \alpha = \frac12\cdot\buildrel\smile\over(AB)\).
Theorem on arcs subtended by equal chords
Equal chords subtend equal arcs smaller than semicircles.
And vice versa: equal arcs are subtended by equal chords.
Proof
1) Let \(AB=CD\) . Let us prove that the smaller semicircles of the arc .
On three sides, therefore, \(\angle AOB=\angle COD\) . But because \(\angle AOB, \angle COD\) - central angles supported by arcs \(\buildrel\smile\over(AB), \buildrel\smile\over(CD)\) accordingly, then \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\).
2) If \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\), That \(\triangle AOB=\triangle COD\) on two sides \(AO=BO=CO=DO\) and the angle between them \(\angle AOB=\angle COD\) . Therefore, and \(AB=CD\) .
Theorem
If the radius bisects the chord, then it is perpendicular to it.
The converse is also true: if the radius is perpendicular to the chord, then at the point of intersection it bisects it.
Proof
1) Let \(AN=NB\) . Let us prove that \(OQ\perp AB\) .
Consider \(\triangle AOB\) : it is isosceles, because \(OA=OB\) – radii of the circle. Because \(ON\) is the median drawn to the base, then it is also the height, therefore, \(ON\perp AB\) .
2) Let \(OQ\perp AB\) . Let us prove that \(AN=NB\) .
Similarly, \(\triangle AOB\) is isosceles, \(ON\) is the height, therefore, \(ON\) is the median. Therefore, \(AN=NB\) .
\[(\Large(\text(Theorems related to the lengths of segments)))\]
Theorem on the product of chord segments
If two chords of a circle intersect, then the product of the segments of one chord is equal to the product of the segments of the other chord.
Proof
Let the chords \(AB\) and \(CD\) intersect at the point \(E\) .
Consider the triangles \(ADE\) and \(CBE\) . In these triangles, angles \(1\) and \(2\) are equal, since they are inscribed and rest on the same arc \(BD\), and angles \(3\) and \(4\) are equal as vertical. Triangles \(ADE\) and \(CBE\) are similar (based on the first criterion of similarity of triangles).
Then \(\dfrac(AE)(EC) = \dfrac(DE)(BE)\), from which \(AE\cdot BE = CE\cdot DE\) .
Tangent and secant theorem
The square of a tangent segment is equal to the product of a secant and its outer part.
Proof
Let the tangent pass through the point \(M\) and touch the circle at the point \(A\) . Let the secant pass through the point \(M\) and intersect the circle at the points \(B\) and \(C\) so that \(MB< MC\) . Покажем, что \(MB\cdot MC = MA^2\) .
Consider the triangles \(MBA\) and \(MCA\) : \(\angle M\) is common, \(\angle BCA = 0.5\cdot\buildrel\smile\over(AB)\). According to the theorem about the angle between a tangent and a secant, \(\angle BAM = 0.5\cdot\buildrel\smile\over(AB) = \angle BCA\). Thus, triangles \(MBA\) and \(MCA\) are similar at two angles.
From the similarity of triangles \(MBA\) and \(MCA\) we have: \(\dfrac(MB)(MA) = \dfrac(MA)(MC)\), which is equivalent to \(MB\cdot MC = MA^2\) .
Consequence
The product of a secant drawn from the point \(O\) by its external part does not depend on the choice of the secant drawn from the point \(O\) .
