Circle equation. Equation of a circle and a straight line Show that this equation defines a circle online

Class: 8

The purpose of the lesson: introduce the equation of a circle, teach students to draw up an equation of a circle according to a finished drawing, build a circle according to a given equation.

Equipment: interactive board.

Lesson plan:

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

During the classes

2. Repetition:

− (Appendix 1 slide 2) write down the formula for finding the coordinates of the middle of the 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 the equations of a circle centered at a point ( a;b) and centered at the origin.

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

X 2 + y 2 = R 2 is the equation of a circle centered at the origin.

(Slide 7)

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

• 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 circle equation.

4. Problem solving.

In tasks No. 1 - No. 6, draw up the equations of the circle according to the finished drawings.

(Slide 14)

№ 7. Fill in the table.

(Slide 15)

№ 8. Construct circles in the 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 is the diameter of the circle.

Given:		Decision:
		R	Center coordinates
1	BUT(0 ; -6) AT(0 ; 2)	AB 2 = (0 – 0) 2 + (2 + 6) 2 ; AB 2 = 64; AB = 8 .	BUT(0; -6) AT(0 ; 2) With(0 ; – 2) – Centre
2	BUT(-2 ; 0) AT(4 ; 0)	AB 2 = (4 + 2) 2 + (0 + 0) 2 ; AB 2 = 36; AB = 6.	BUT (-2;0) AT (4 ;0) With(1 ; 0) – Centre

(Slide 17)

№ 10. Write the equation of a circle centered at the origin passing through the point To(-12;5).

Decision.

R2 = OK 2 = (0 + 12) 2 + (0 – 5) 2 = 144 + 25 = 169;
R= 13;

Circle equation: x 2 + y 2 = 169 .

(Slide 18)

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

Decision.

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

Circle equation: ( X - 3) 2 + (y + 1) 2 = 10.

(Slide 19)

№ 12. Write the equation of a circle with a center BUT(3;2) passing through AT(7;5).

Decision.

1. The center of the circle - BUT(3;2);
2.R = AB;
AB 2 = (7 – 3) 2 + (5 – 2) 2 = 25; AB = 5;
3. Circle equation ( X – 3) 2 + (at − 2) 2 = 25.

(Slide 20)

№ 13. Check if points lie BUT(1; -1), AT(0;8), With(-3; -1) on the circle given by the equation ( X + 3) 2 + (at − 4) 2 = 25.

Decision.

I. Substitute the coordinates of the point BUT(1; -1) into the circle equation:

(1 + 3) 2 + (−1 − 4) 2 = 25;
4 2 + (−5) 2 = 25;
16 + 25 = 25;
41 \u003d 25 - equality is incorrect, which means BUT(1; -1) does not lie on the circle given by the equation ( X + 3) 2 + (at − 4) 2 = 25.

II. Substitute the coordinates of the point AT(0;8) into the circle equation:

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

III. Substitute the coordinates of the point With(-3; -1) into the circle equation:

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

Summary of the lesson.

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

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 definition of a circle

Equality (1) is circle equation radius R centered 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 М( X; at) is an arbitrary point of this circle.

Since |CM| = \(\sqrt((x - a)^2 + (y - b)^2) \), 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 the general equation of a circle or the equation of a circle of radius R centered at the point ( a; b). For example, the equation

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

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

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

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

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

Task 1. Write the equation for a circle of radius R = 7 centered at the origin.

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

x 2 + at 2 = 49.

Task 2. Write the equation for a circle of radius R = 9 centered 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 circle equation (2), we see that a = -3, b= 5, R = 10. Therefore, С(-3; 5), R = 10.

Task 4. Prove that the equation

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

is the circle equation. 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 centered at the point C(-1; -1) touching the straight line AB if A (2; -1), B(-1; 3).

Let's write the equation of the straight line AB:

or 4 X + 3y-5 = 0.

Since the circle is tangent to the 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 the point C (-1; -1) - the center of the circle to the 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 the positive direction of the O axis X, then the abscissa and ordinate of the 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 centered 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 the canonical equation for 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

Lesson topic: Circle equation

Lesson Objectives:

Educational: Derive the circle equation, considering the solution of this problem as one of the possibilities of applying the coordinate method.

Be able to:

– Recognize the equation of a circle according to the proposed equation, teach students to draw up an equation of a circle according to a finished drawing, build a circle according to a given equation.

Educational : Formation of critical thinking.

Educational : Development of the ability to make algorithmic prescriptions and the ability to act in accordance with the proposed algorithm.

Be able to:

– See the problem and plan ways to solve it.

– Summarize your thoughts orally and in writing.

Lesson type: assimilation of new knowledge.

Equipment : PC, multimedia projector, screen.

Lesson plan:

1. Opening speech - 3 min.

2. Updating knowledge - 2 min.

3. Statement of the problem and its solution -10 min.

4. Frontal fastening of the new material - 7 min.

5. Independent work in groups - 15 min.

6. Presentation of the work: discussion - 5 min.

7. The result of the lesson. Homework - 3 min.

During the classes

The purpose of this stage: Psychological mood of students; Involvement of all students in the learning process, creating a situation of success.

1. Organizing time.

3 minutes

Guys! You met the circle back in the 5th and 8th grades. What do you know about her?

You know a lot, and this data can be used in solving geometric problems. But for solving problems in which the coordinate method is used, this is not enough.Why?

Absolutely right.

Therefore, the main goal of today's lesson is to derive the equation of a circle from the geometric properties of a given line and apply it to solve geometric problems.

Let it gomotto of the lesson the words of the Central Asian scientist-encyclopedist Al-Biruni will become: “Knowledge is the most excellent of possessions. Everyone strives for it, but it does not come by itself.”

Write the topic of the lesson in a notebook.

Definition of a circle.

Radius.

Diameter.

Chord. Etc.

We do not yet know the general form of the circle equation.

Students list everything they know about the circle.

slide 2

slide 3

The purpose of the stage is to get an idea of ​​the quality of learning by students of the material, to determine the basic knowledge.

2. Knowledge update.

2 minutes

When deriving the circle equation you will need the already known definition of a circle and a formula that allows you to find the distance between two points by their coordinates.Let's remember these facts /Prepetition of material previously studied/:

– Write down the formula for finding the coordinates of the midpoint of a segment.

– Write down the formula for calculating the length of a vector.

– Write down the formula for finding the distance between points (length of the segment).

Editing records...

Geometric workout.

Given pointsA (-1; 7) andIn (7; 1).

Calculate the coordinates of the midpoint of the segment AB and its length.

Checks the correctness of execution, corrects calculations ...

One student at the blackboard, and the rest write down formulas in notebooks

A circle is a geometric figure consisting of all points located at a given distance from a given point.

| AB | \u003d √ (x - x) ² + (y - y) ²

M(x;y), A(x;y)

Calculate: C (3; 4)

| AB | = 10

With lay 4

slide 5

3. Formation of new knowledge.

12 minutes

Purpose: the formation of the concept - the equation of the circle.

Solve the problem:

A circle with center A(x; y) is constructed in a rectangular coordinate system. M(x; y) - arbitrary point of the circle. Find the radius of the circle.

Will the coordinates of any other point satisfy this equality? Why?

Let's square both sides of the equation.As a result, we have:

r² \u003d (x - x) ² + (y - y) ² is the equation of the circle, where (x; y) is the coordinates of the center of the circle, (x; y) is the coordinates of an arbitrary point lying on the circle, r is the radius of the circle.

Solve the problem:

What will be the equation of a circle centered at the origin?

So, what do you need to know to write the equation of a circle?

Suggest an algorithm for compiling the circle equation.

Conclusion: ... write in a notebook.

A radius is a segment connecting the center of a circle with an arbitrary point lying on the circle. Therefore, r \u003d | AM | \u003d √ (x - x)² + (y - y)²

Any point on a circle lies on that circle.

Students write in notebooks.

(0;0)-coordinates of the center of the circle.

x² + y² = r², where r is the radius of the circle.

The coordinates of the center of the circle, the radius, any point on the circle...

They propose an algorithm...

Write down the algorithm in a notebook.

slide 6

Slide 7

Slide 8

The teacher writes the equation on the blackboard.

Slide 9

4. Primary fastening.

23 minutes

Target:reproduction by students of the material that has just been perceived to prevent the loss of the formed ideas and concepts. Consolidation of new knowledge, ideas, concepts based on theirapplications.

ZUN control

Let's apply the acquired knowledge in solving the following problems.

Task: From the proposed equations, name the numbers of those that are the equations of the circle. And if the equation is the equation of a circle, then name the coordinates of the center and indicate the radius.

Not every equation of the second degree with two variables defines a circle.

4x² + y² \u003d 4-ellipse equation.

x²+y²=0-dot.

x² + y² \u003d -4-this equation does not define any figure.

Guys! What do you need to know to write an equation for a circle?

Solve the problem No. 966 p. 245 (textbook).

The teacher calls the student to the blackboard.

Is the data specified in the condition of the problem enough to draw up an equation for a circle?

Task:

Write the equation for a circle centered at the origin and with a diameter of 8.

Task : draws a circle.

Center has coordinates?

Determine the radius... and build

Task on page 243 (textbook) is understood orally.

Using the problem solving plan from p.243, solve the problem:

Write the equation of a circle centered at point A(3;2) if the circle passes through point B(7;5).

1) (x-5) ² + (y-3) ² \u003d 36 - circle equation; (5; 3), r \u003d 6.

2) (x-1)² + y² \u003d 49 - circle equation; (1; 0), r \u003d 7.

3) x² + y² \u003d 7 - circle equation; (0; 0), r \u003d √7.

4) (x + 3)² + (y-8)² \u003d 2- circle equation; (-3;8),r=√2.

5) 4x² + y² \u003d 4 is not an equation of a circle.

6) x² + y² = 0- is not an equation of a circle.

7) x² + y² = -4- is not an equation of a circle.

Know the coordinates of the center of the circle.

Radius length.

Substitute the coordinates of the center and the length of the radius into the general equation of a circle.

Solve problem No. 966 p. 245 (textbook).

Enough data.

They solve the problem.

Since the diameter of a circle is twice its radius, then r=8÷2=4. Therefore, x² + y² = 16.

Perform the construction of circles

Textbook work. Task on page 243.

Given: A (3; 2) - the center of the circle; В(7;5)є(А;r)

Find: circle equation

Solution: r² \u003d (x - x)² + (y - y)²

r² \u003d (x -3)² + (y -2)²

r = AB, r² = AB²

r² =(7-3)²+(5-2)²

r²=25

(x -3)² + (y -2)² \u003d 25

Answer: (x -3)² + (y -2)² \u003d 25

slide 10-13

Solving typical problems by pronouncing the solution in a loud speech.

The teacher calls one student to write down the resulting equation.

Return to slide 9

Discussion of a plan for solving this problem.

Slide. fifteen. The teacher calls one student to the board to solve this problem.

slide 16.

slide 17.

5. Summary of the lesson.

5 minutes

Reflection of activities in the classroom.

Homework: §3, item 91, control questions No. 16,17.

Problems No. 959(b, d, e), 967.

Task for additional assessment (problem task): Construct a circle given by the equation

x² + 2x + y² -4y = 4.

What did we talk about in class?

What did you want to receive?

What was the purpose of the lesson?

What tasks can be solved by our "discovery"?

Which of you believes that you have achieved the goal set by the teacher in the lesson by 100%, by 50%; didn't reach the goal...?

Grading.

Write down homework.

Students answer questions posed by the teacher. Conduct a self-assessment of their own performance.

Students need to express in a word the result and ways to achieve it.

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 the line $L$ if this equation is satisfied by the coordinates of any point belonging to the line $L$ and not satisfied by any point not belonging to the line $L.$

Circle equation

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

Figure 2. Circle in Cartesian coordinates

The distance from the center of the circle to the 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 centered at the point $(x_0,y_0)$ and radius $r$.

In particular, if the center of the circle coincides with the origin. Then the equation of the 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 the points $A$ and $B$ have the coordinates $\left\(x_1,\ y_1\right\)$ and $\(x_2,\ y_2\)$, respectively, and the points $A$ and $B$ are chosen so that that the line $l$ is the perpendicular bisector to the segment $AB$. We choose an arbitrary point $M=\(x,y\)$ belonging to the line $l$ (Fig. 3).

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

Find the lengths of these sides using the formula for the distance between points:

Hence

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 get that the equation of a straight line in the Cartesian coordinate system has the following form:

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

Example 1

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

Decision.

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

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

Find the radius of the circle as the distance from the point $(2,\ 4)$ to the point $(0,0)$

We get the equation of the circle has the form:

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

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