Sinusitis

Equation of a circle in general form. Equation of a circle

Objective of the lesson: introduce the equation of a circle, teach students to compose an equation of a circle using a ready-made drawing, and construct a circle using a given equation.

Equipment: interactive whiteboard.

Lesson plan:

Organizational moment – 3 min.
Repetition. Organization of mental activity – 7 min.
Explanation of new material. Derivation of the equation of a circle – 10 min.
Consolidation of the studied material – 20 min.
Lesson summary – 5 min.

Lesson progress

2. Repetition:

− (Appendix 1 Slide 2) write down the formula for finding the coordinates of the middle of a segment;

− (Slide 3) Z Write the formula for the distance between points (the length of the segment).

3. Explanation of new material.

(Slides 4 – 6) Define the equation of a circle. Derive equations of a circle with center at point ( A;b) and centered at the origin.

(X – A ) 2 + (at – b ) 2 = R 2 – equation of a circle with center WITH (A;b) , radius R , X And at – coordinates of an arbitrary point on the circle .

X 2 + y 2 = R 2 – equation of a circle with center at the origin.

(Slide 7)

In order to create the equation of a circle, you need to:

know the coordinates of the center;
know the length of the radius;
Substitute the coordinates of the center and the length of the radius into the equation of the circle.

4. Problem solving.

In tasks No. 1 – No. 6, compose equations of a circle using ready-made drawings.

(Slide 14)

№ 7. Fill out the table.

(Slide 15)

№ 8. Construct circles in your notebook given by the equations:

A) ( X – 5) 2 + (at + 3) 2 = 36;
b) (X + 1) 2 + (at– 7) 2 = 7 2 .

(Slide 16)

№ 9. Find the coordinates of the center and the length of the radius if AB– diameter of the circle.

Given:		Solution:
Given:		R	Center coordinates
1	A(0 ; -6) IN(0 ; 2)	AB 2 = (0 – 0) 2 + (2 + 6) 2 ; AB 2 = 64; AB = 8 .	A(0; -6) IN(0 ; 2) WITH(0 ; – 2) – center
2	A(-2 ; 0) IN(4 ; 0)	AB 2 = (4 + 2) 2 + (0 + 0) 2 ; AB 2 = 36; AB = 6.	A (-2;0) IN (4 ;0) WITH(1 ; 0) – center

(Slide 17)

№ 10. Write an equation for a circle with center at the origin and passing through the point TO(-12;5).

Solution.

R 2 = OK 2 = (0 + 12) 2 + (0 – 5) 2 = 144 + 25 = 169;
R= 13;

Equation of a circle: x 2 + y 2 = 169 .

(Slide 18)

№ 11. Write an equation for a circle passing through the origin and centered at WITH(3; - 1).

Solution.

R2= OS 2 = (3 – 0) 2 + (–1–0) 2 = 9 + 1 = 10;

Equation of a circle: ( X - 3) 2 + (y + 1) 2 = 10.

(Slide 19)

№ 12. Write an equation for a circle with a center A(3;2), passing through IN(7;5).

Solution.

1. Center of the circle – A(3;2);
2.R = AB;
AB 2 = (7 – 3) 2 + (5 – 2) 2 = 25; AB = 5;
3. Equation of a circle ( X – 3) 2 + (at − 2) 2 = 25.

(Slide 20)

№ 13. Check if the points lie A(1; -1), IN(0;8), WITH(-3; -1) on the circle defined by the equation ( X + 3) 2 + (at − 4) 2 = 25.

Solution.

I. Let's substitute the coordinates of the point A(1; -1) into the equation of a circle:

(1 + 3) 2 + (−1 − 4) 2 = 25;
4 2 + (−5) 2 = 25;
16 + 25 = 25;
41 = 25 – the equality is false, which means A(1; -1) doesn't lie on the circle given by the equation ( X + 3) 2 + (at − 4) 2 = 25.

II. Let's substitute the coordinates of the point IN(0;8) into the equation of a circle:

(0 + 3) 2 + (8 − 4) 2 = 25;
3 2 + 4 2 = 25;
9 + 16 = 25;
IN(0;8)lies X + 3) 2 + (at − 4) 2 = 25.

III. Let's substitute the coordinates of the point WITH(-3; -1) into the equation of a circle:

(−3 + 3) 2 + (−1− 4) 2 = 25;
0 2 + (−5) 2 = 25;
25 = 25 – the equality is true, which means WITH(-3; -1) lies on the circle given by the equation ( X + 3) 2 + (at − 4) 2 = 25.

Lesson summary.

Repeat: equation of a circle, equation of a circle with its center at the origin.
(Slide 21) Homework.

Equation of a line on a plane

Let us first introduce the concept of the equation of a line in a two-dimensional coordinate system. Let an arbitrary line $L$ be constructed in the Cartesian coordinate system (Fig. 1).

Figure 1. Arbitrary line in the coordinate system

Definition 1

An equation with two variables $x$ and $y$ is called an equation of line $L$ if this equation is satisfied by the coordinates of any point belonging to line $L$ and not satisfied by any point not belonging to line $L.$

Equation of a circle

Let us derive the equation of a circle in the Cartesian coordinate system $xOy$. Let the center of the circle $C$ have coordinates $(x_0,y_0)$, and the radius of the circle be equal to $r$. Let point $M$ with coordinates $(x,y)$ be an arbitrary point of this circle (Fig. 2).

Figure 2. Circle in Cartesian coordinate system

The distance from the center of the circle to point $M$ is calculated as follows

But, since $M$ lies on the circle, we get $CM=r$. Then we get the following

Equation (1) is the equation of a circle with center at point $(x_0,y_0)$ and radius $r$.

In particular, if the center of the circle coincides with the origin. That equation of a circle has the form

Equation of a straight line.

Let us derive the equation of the straight line $l$ in the Cartesian coordinate system $xOy$. Let points $A$ and $B$ have coordinates $\left$x_1,\ y_1\right$$ and $$x_2,\ y_2$$, respectively, and points $A$ and $B$ are chosen so that that the line $l$ is the perpendicular bisector of the segment $AB$. Let us choose an arbitrary point $M=$x,y$$ belonging to the straight line $l$ (Fig. 3).

Since the line $l$ is the perpendicular bisector to the segment $AB$, then the point $M$ is equidistant from the ends of this segment, that is, $AM=BM$.

Let's find the lengths of these sides using the formula for the distance between points:

Hence

Let us denote by $a=2\left(x_1-x_2\right),\ b=2\left(y_1-y_2\right),\ c=(x_2)^2+(y_2)^2-(x_1)^2 -(y_1)^2$, We find that the equation of a straight line in a Cartesian coordinate system has the following form:

An example of a problem on finding the equations of lines in a Cartesian coordinate system

Example 1

Find the equation of a circle with center at point $(2,\ 4)$. Passing through the origin of coordinates and a straight line parallel to the $Ox,$ axis passing through its center.

Solution.

Let's first find the equation of this circle. To do this, we will use the general equation of a circle (derived above). Since the center of the circle lies at the point $(2,\ 4)$, we get

\[((x-2))^2+((y-4))^2=r^2\]

Let's find the radius of the circle as the distance from the point $(2,\ 4)$ to the point $(0,0)$

We find that the equation of a circle has the form:

\[((x-2))^2+((y-4))^2=20\]

Let us now find the equation of a circle using special case 1. We obtain

Circumference is the set of points in the plane equidistant from a given point called the center.

If point C is the center of the circle, R is its radius, and M is an arbitrary point on the circle, then by the definition of a circle

Equality (1) is equation of a circle radius R with center at point C.

Let a rectangular cartesian system coordinates (Fig. 104) and point C( A; b) is the center of a circle of radius R. Let M( X; at) is an arbitrary point of this circle.

Since |SM| = $\sqrt((x - a)^2 + (y - b)^2) $, then equation (1) can be written as follows:

$\sqrt((x - a)^2 + (y - b)^2) $ = R

(x-a) 2 + (y - b) 2 = R 2 (2)

Equation (2) is called general equation of a circle or the equation of a circle of radius R with center at point ( A; b). For example, the equation

(x - l) 2 + ( y + 3) 2 = 25

is the equation of a circle of radius R = 5 with center at point (1; -3).

If the center of the circle coincides with the origin of coordinates, then equation (2) takes the form

x 2 + at 2 = R 2 . (3)

Equation (3) is called canonical equation of a circle .

Task 1. Write the equation of a circle of radius R = 7 with its center at the origin.

By directly substituting the radius value into equation (3) we obtain

x 2 + at 2 = 49.

Task 2. Write the equation of a circle of radius R = 9 with center at point C(3; -6).

Substituting the value of the coordinates of point C and the value of the radius into formula (2), we obtain

(X - 3) 2 + (at- (-6)) 2 = 81 or ( X - 3) 2 + (at + 6) 2 = 81.

Task 3. Find the center and radius of a circle

(X + 3) 2 + (at-5) 2 =100.

Comparing this equation with the general equation of a circle (2), we see that A = -3, b= 5, R = 10. Therefore, C(-3; 5), R = 10.

Task 4. Prove that the equation

x 2 + at 2 + 4X - 2y - 4 = 0

is the equation of a circle. Find its center and radius.

Let's transform left side of this equation:

x 2 + 4X + 4- 4 + at 2 - 2at +1-1-4 = 0

(X + 2) 2 + (at - 1) 2 = 9.

This equation is the equation of a circle centered at (-2; 1); The radius of the circle is 3.

Task 5. Write the equation of a circle with center at point C(-1; -1) tangent to line AB, if A (2; -1), B(- 1; 3).

Let's write the equation of line AB:

or 4 X + 3y-5 = 0.

Since a circle touches a given line, the radius drawn to the point of contact is perpendicular to this line. To find the radius, you need to find the distance from point C(-1; -1) - the center of the circle to straight line 4 X + 3y-5 = 0:

Let's write the equation of the desired circle

(x +1) 2 + (y +1) 2 = 144 / 25

Let a circle be given in a rectangular coordinate system x 2 + at 2 = R 2 . Consider its arbitrary point M( X; at) (Fig. 105).

Let the radius vector OM> point M forms an angle of magnitude t with positive direction of the O axis X, then the abscissa and ordinate of point M change depending on t

(0 t x and y through t, we find

x= Rcos t ; y= R sin t , 0 t

Equations (4) are called parametric equations of a circle with center at the origin.

Task 6. The circle is given by the equations

x= $\sqrt(3)$cos t, y= $\sqrt(3)$sin t, 0 t

Write down the canonical equation of this circle.

It follows from the condition x 2 = 3 cos 2 t, at 2 = 3 sin 2 t. Adding these equalities term by term, we get

x 2 + at 2 = 3(cos 2 t+ sin 2 t)

or x 2 + at 2 = 3