How to solve the equation of a circle. Circle on the coordinate plane

Definition 1. Number axis ( number line, coordinate line) Ox is the straight line on which point O is selected origin (origin of coordinates)(Fig. 1), direction

O → x

listed as positive direction and a segment is marked, the length of which is taken to be unit of length.

Definition 2. A segment whose length is taken as a unit of length is called scale.

Each point on the number axis has a coordinate that is a real number. The coordinate of point O is zero. The coordinate of an arbitrary point A lying on the ray Ox is equal to the length of the segment OA. The coordinate of an arbitrary point A of the numerical axis that does not lie on the ray Ox is negative, and in absolute value is equal to the length of the segment OA.

Definition 3. Rectangular Cartesian coordinate system Oxy on the plane call two mutually perpendicular numerical axes Ox and Oy with the same scale And common reference point at point O, and such that the rotation from ray Ox at an angle of 90° to ray Oy is carried out in the direction counterclockwise(Fig. 2).

Note. The rectangular Cartesian coordinate system Oxy, shown in Figure 2, is called right coordinate system, unlike left coordinate systems, in which the rotation of the beam Ox at an angle of 90° to the beam Oy is carried out in a clockwise direction. In this guide we we consider only right-handed coordinate systems, without specifically specifying it.

If we introduce some system of rectangular Cartesian coordinates Oxy on the plane, then each point of the plane will acquire two coordinates – abscissa And ordinate, which are calculated as follows. Let A be an arbitrary point on the plane. Let us drop perpendiculars from point A A.A. 1 and A.A. 2 to straight lines Ox and Oy, respectively (Fig. 3).

Definition 4. The abscissa of point A is the coordinate of the point A 1 on the number axis Ox, the ordinate of point A is the coordinate of the point A 2 on the number axis Oy.

Designation Coordinates (abscissa and ordinate) of the point A in the rectangular Cartesian coordinate system Oxy (Fig. 4) is usually denoted A(x;y) or A = (x; y).

Note. Point O, called origin, has coordinates O(0 ; 0) .

Definition 5. In the rectangular Cartesian coordinate system Oxy, the numerical axis Ox is called the abscissa axis, and the numerical axis Oy is called the ordinate axis (Fig. 5).

Definition 6. Each rectangular Cartesian coordinate system divides the plane into 4 quarters (quadrants), the numbering of which is shown in Figure 5.

Definition 7. The plane on which a rectangular Cartesian coordinate system is given is called coordinate plane.

Note. The abscissa axis is specified on the coordinate plane by the equation y= 0, the ordinate axis is given on the coordinate plane by the equation x = 0.

Statement 1. Distance between two points coordinate plane

A 1 (x 1 ;y 1) And A 2 (x 2 ;y 2)

calculated according to the formula

Proof . Consider Figure 6.

If you place the unit number circle on the coordinate plane, then you can find the coordinates for its points. The number circle is positioned so that its center coincides with the origin of the plane, i.e., point O (0; 0).

Usually on the unit number circle the points corresponding to the origin of the circle are marked

• quarters - 0 or 2π, π/2, π, (2π)/3,
• middle quarters - π/4, (3π)/4, (5π)/4, (7π)/4,
• thirds of quarters - π/6, π/3, (2π)/3, (5π)/6, (7π)/6, (4π)/3, (5π)/3, (11π)/6.

On the coordinate plane, with the above location of the unit circle on it, you can find the coordinates corresponding to these points of the circle.

The coordinates of the ends of the quarters are very easy to find. At point 0 of the circle, the x coordinate is 1, and the y coordinate is 0. We can denote it as A (0) = A (1; 0).

The end of the first quarter will be located on the positive y-axis. Therefore, B (π/2) = B (0; 1).

The end of the second quarter is on the negative semi-axis: C (π) = C (-1; 0).

End of third quarter: D ((2π)/3) = D (0; -1).

But how to find the coordinates of the midpoints of the quarters? For this they build right triangle. Its hypotenuse is a segment from the center of the circle (or origin) to the midpoint of the quarter circle. This is the radius of the circle. Since there is a unit circle, the hypotenuse is equal to 1. Next, draw a perpendicular from a point on the circle to any axis. Let it be towards the x axis. The result is a right triangle, the lengths of the legs of which are the x and y coordinates of the point on the circle.

A quarter circle is 90º. And half a quarter is 45º. Since the hypotenuse is drawn to the midpoint of the quadrant, the angle between the hypotenuse and the leg extending from the origin is 45º. But the sum of the angles of any triangle is 180º. Consequently, the angle between the hypotenuse and the other leg also remains 45º. This results in an isosceles right triangle.

From the Pythagorean theorem we obtain the equation x 2 + y 2 = 1 2. Since x = y and 1 2 = 1, the equation simplifies to x 2 + x 2 = 1. Solving it, we get x = √½ = 1/√2 = √2/2.

Thus, the coordinates of the point M 1 (π/4) = M 1 (√2/2; √2/2).

In the coordinates of the points of the midpoints of the other quarters, only the signs will change, and the modules of the values ​​will remain the same, since the right triangle will only be turned over. We get:
M 2 ((3π)/4) = M 2 (-√2/2; √2/2)
M 3 ((5π)/4) = M 3 (-√2/2; -√2/2)
M 4 ((7π)/4) = M 4 (√2/2; -√2/2)

When determining the coordinates of the third parts of the quarters of a circle, a right triangle is also constructed. If we take the point π/6 and draw a perpendicular to the x-axis, then the angle between the hypotenuse and the leg lying on the x-axis will be 30º. It is known that a leg lying opposite an angle of 30º is equal to half the hypotenuse. This means that we have found the y coordinate, it is equal to ½.

Knowing the lengths of the hypotenuse and one of the legs, using the Pythagorean theorem we find the other leg:
x 2 + (½) 2 = 1 2
x 2 = 1 - ¼ = ¾
x = √3/2

Thus T 1 (π/6) = T 1 (√3/2; ½).

For the point of the second third of the first quarter (π/3), it is better to draw a perpendicular to the axis to the y axis. Then the angle at the origin will also be 30º. Here the x coordinate will be equal to ½, and y, respectively, √3/2: T 2 (π/3) = T 2 (½; √3/2).

For other points of the third quarters, the signs and order of the coordinate values ​​will change. All points that are closer to the x axis will have a modulus x coordinate value equal to √3/2. Those points that are closer to the y axis will have a modulus y value equal to √3/2.
T 3 ((2π)/3) = T 3 (-½; √3/2)
T 4 ((5π)/6) = T 4 (-√3/2; ½)
T 5 ((7π)/6) = T 5 (-√3/2; -½)
T 6 ((4π)/3) = T 6 (-½; -√3/2)
T 7 ((5π)/3) = T 7 (½; -√3/2)
T 8 ((11π)/6) = T 8 (√3/2; -½)

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 coordinate system (Fig. 104) and a 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 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 the 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 canonical equation 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

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:

1. Organizational moment – ​​3 min.
2. Repetition. Organization mental activity– 7 min.
3. Explanation of new material. Derivation of the equation of a circle – 10 min.
4. Consolidation of the studied material – 20 min.
5. 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:
		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.

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