Lesson topic: “Mutual arrangement of two circles. Mutual arrangement of two circles on a plane Mutual arrangement of two circles

Let the circles be given by a vector from the origin to the center and the radius of this circle.

Consider circles A and B with radii Ra and Rb and radius vectors (vector to the center) a and b. Moreover, Oa and Ob are their centers. Without loss of generality of reasoning, we will assume that Ra> Rb.

Then the following conditions are satisfied:

Intersection points of two circles

Suppose A and B meet at two points. Let's find these intersection points.

To do this, the vector from a to point P, which lies on the circle A and lies on OaOb. To do this, you need to take the vector b - a, which will be the vector between the two centers, normalize (replace with a co-directional unit vector) and multiply by Ra. The resulting vector will be denoted as p. You can see this configuration in fig. 6

Rice. 6. Vectors a, b, p and where they live.

Let us denote i1 and i2 as vectors from a to the intersection points I1 and I2 of two circles. It becomes obvious that i1 and i2 are obtained by rotation from p. Because we know all sides of triangles OaI1Ob and OaI2Ob (Radii and distance between centers), we can get this angle fi, rotation by which vector p in one direction will give I1, and in the other I2.

By the cosine theorem, it is equal to:

If you turn p to fi, you get either i1 or i2, depending on which direction you turn. Further, the vector i1 or i2 must be added to a to obtain the intersection point

This method will work even if the center of one circle is inside the other. But there exactly the vector p will have to be set in the direction from a to b, which we did. If you build p based on another circle, then nothing will come of it

Well, in conclusion, one fact should be mentioned to everything: if the circles touch, then it is easy to make sure that P is the point of tangency (this is true for both internal and external tangencies).
Here you can see the visualization (click to launch).

This method is working, but instead of the rotation angle, you can calculate its cosine, and through it the sine, and then use them when you rotate the vector. This will greatly simplify calculations by eliminating trigonometric functions in your code.

Class 7Ж, З

Lesson topic: "The relative position of two circles"
Purpose: to know the possible cases of mutual arrangement of two circles; apply knowledge in solving problems.

Tasks: Educational: to promote the creation and consolidation of students' visual representation of the possible cases of the location of two circles, students will be able to:

Establish a relationship between the relative position of the circles, their radii and the distance between their centers;

Analyze the geometric structure and mentally modify it,

Develop planimetric imagination.

Students will be able to apply theoretical knowledge to problem solving.

Lesson type: lesson introduction and consolidation of new knowledge of the material.

Equipment: presentation for the lesson; compasses, ruler, pencil and textbook for each student.

Tutorial: . "Geometry Grade 7", Almaty "Atamura" 2012

During the classes.

Organizing time. Homework check.

3. Updating basic knowledge.

Repeat definitions of circle, circle, radius, diameter, chord, distance from point to line.

1) 1) What cases of a line and a circle are known to you?

2) Which line is called a tangent line?

3) Which line is called a secant?

4) The theorem on the diameter perpendicular to the chord?

5) How is the tangent going in relation to the radius of the circle?

6) Fill in the table (on cards).

Students, under the guidance of a teacher, solve and analyze problems.

1) Line a - tangent to a circle with center O. Point A is given on line a. The angle between the tangent and the segment OA is 300. Find the length of the segment OA if the radius is 2.5 m.

2) Determine the relative position of the straight line and the circle if:

1.R = 16cm, d = 12cm 2.R = 5cm, d = 4.2cm 3.R = 7.2dm, d = 3.7dm 4. R = 8cm, d = 1.2dm 5. R = 5 cm, d = 50mm

a) the line and the circle have no common points;

b) the straight line is tangent to the circle;

c) the straight line intersects the circle.

d is the distance from the center of the circle to the straight line, R is the radius of the circle.

3) What can be said about the relative position of the line and the circle if the diameter of the circle is 10.3 cm, and the distance from the center of the circle to the line is 4.15 cm; 2 dm; 103 mm; 5.15 cm, 1 dm 3 cm.

4) Given a circle with a center O and point A. Where is point A, if the radius of the circle is 7 cm, and the length of the segment OA is equal to: a) 4 cm; b) 10 cm; c) 70 mm.

4. Together with the students, find out the topic of the lesson, formulate the goals of the lesson.

5. Introduction of new material.

Practical work in groups.

Construct 3 circles. To each circle, construct one more circle, so that 1) 2 circles do not intersect, 2) 2 circles touch, 3) two circles intersect. Find the radius of each circle and the distance between the centers of the circles, compare the results. What conclusion can be drawn?
2) Summarize and write down in a notebook, cases of mutual arrangement of two circles.

The relative position of two circles on the plane.

The circles do not have common points (do not intersect). (R1 and R2 are the radii of the circles)

If R1 + R2< d,

d - Distance between the centers of the circles.

c) The circles have two points in common. (intersect).

If R1 + R2> d,

Question. Can two circles have three points in common?

6. Consolidation of the studied material.

Look for an error in the data or in the statement and correct it based on your opinion:
A) Two circles touch. Their radii are equal to R = 8 cm and r = 2 cm, the distance between the centers is d = 6.
B) Two circles have at least two common points.
B) R = 4, r = 3, d = 5. The circles have no common points.
D) R = 8, r = 6, d = 4. The smaller circle is located inside the larger one.
E) Two circles cannot be located so that one is inside the other.

7. Lesson summary. What have you learned in the lesson? What regularity have you established?

How can two circles be located? When do the circles have one point in common? What is the common point of two circles called? What touches do you know? When do the circles intersect? Which circles are called concentric?

Lesson topic: " Mutual arrangement of two circles on a plane ”.

Target :

Educational - assimilation of new knowledge about the relative position of two circles, preparation for the test

Developing - development of computational skills, development of logical-structural thinking; developing skills for finding rational solutions and achieving final results; development of cognitive activity and creative thinking.

Educational – the formation of students' responsibility, consistency; development of cognitive and aesthetic qualities; the formation of the information culture of students.

Correctional - develop spatial thinking, memory, hand motor skills.

Lesson type: study of new educational material, consolidation.

Lesson type: mixed lesson.

Teaching method: verbal, visual, practical.

Form of study: collective.

Means of education: board

DURING THE CLASSES:

1. Organizational stage

- greetings;

- checking the readiness for the lesson;

2. Updating basic knowledge.
What topics did we cover in the previous lessons?

General view of the equation of the circle?

Run orally:

Blitz survey

3. Introduction of new material.

What do you think and what figure we will consider today…. And if there are two of them ??

How can they be located ???

Children show with their hands (neighbors) how circles can be located ( physical education)

Well, what do you think we should consider today ?? Today we should consider the relative position of two circles. And find out what is the distance between the centers depending on the location.

Lesson topic:« The relative position of two circles. Solving problems.»

1. Concentric circles

2. Disjoint circles

3.External touch

4. Intersecting circles

5. Internally touching

So let's conclude

4. Formation of skills and abilities

Look for an error in the data or in the statement and correct it based on your opinion:

A) Two circles touch. Their radii are equal to R = 8 cm and r = 2 cm, the distance between the centers is d = 6.
B) Two circles have at least two common points.

B) R = 4, r = 3, d = 5. The circles have no common points.

D) R = 8, r = 6, d = 4. The smaller circle is located inside the larger one.

E) Two circles cannot be located so that one is inside the other.

5. Consolidation of skills and abilities.

The circles are tangent externally. The radius of the smaller circle is 3 cm. The radius of the larger one is 5 cm. What is the distance between the centers?

Solution: 3 + 5 = 8 (cm)

The circles are tangent inwardly. The radius of the smaller circle is 3 cm. The radius of the larger circle is 5 cm. What is the distance between the centers of the circles?

Solution: 5-3 = 2 (cm)

The circles are tangent inwardly. The distance between the centers of the circles is 2.5 cm. What are the radii of the circles?

answer: (5.5 cm and 3 cm), (6.5 cm and 4 cm), etc.

UNDERSTANDING CHECK

1) How can two circles be located?

2) In what case do the circles have one common point?

3) What is the common point of two circles called?

4) What touches do you know?

5) When do the circles intersect?

6) What circles are called concentric?

Additional tasks on the topic: Vectors. Coordinate method"(If time remains)

1) E (4; 12), F (-4; -10), G (-2; 6), H (4; -2) Find:

a) coordinates of vectors EF, GH

b) the length of the vector FG

c) coordinates of point O - midpoint EF

coordinates of point W - midpoint of GH

d) the equation of a circle with a diameter FG

e) the equation of the straight line FH

6. Homework

& 96 # 1000. Which of these equations are equations of the circle. Find center and radius

7. Lesson summary(3 min.)

(give a qualitative assessment of the work of the class and individual students).

8. Stage of reflection(2 minutes.)

(initiate reflection of students about their emotional state, their activities, interaction with the teacher and classmates using drawings)

Ministry of Education and Science of the Russian Federation

Municipal budgetary educational institution

the city of Novosibirsk "Gymnasium No. 4"

Section: mathematics

RESEARCH

on this topic:

PROPERTIES OF TWO CONTACTING CIRCUITS

Grade 10 students:

Khaziakhmetova Radik Ildarovich

Zubarev Evgeny Vladimirovich

Supervisor:

L.L. Barinova

Mathematic teacher

Highest qualification category

§ 1.Introduction ……… .. …………………………. ………………………………………………… 3

§ 1.1 Mutual arrangement of two circles ……………………… ... ………… ... ……… 3

§ 2 Properties and their proofs ……………………………………… .. …………… .....….… 4

§ 2.1 Property 1 ……………… ... …………………………………… .. ………………… ...….… 4

§ 2.2 Property 2 ………………………………………………… .. ………………… ... ……… 5

§ 2.3 Property 3 ………………………………………………… .. ………………… ... ……… 6

§ 2.4 Property 4 ………………………………………………… .. ………………… ... ……… 6

§ 2.5 Property 5 ………………………………… .. …………………………………… ... ……… 8

§ 2.6 Property 6 ……………………………………………… .. ……………………… ... ……… 9

§ 3 Tasks ………………………………………………… .. ………………… ...… ... ... ……… ..… 11

References ………………………………………………………………. ………… .13

§ 1. Introduction

Many problems involving two tangent circles can be solved in a shorter and simpler manner, knowing some of the properties that will be presented later.

The relative position of two circles

To begin with, let's discuss the possible relative position of the two circles. There can be 4 different cases.

1.Circles may not intersect.

2. Overlap.

3. Touch at one point outside.

4. Touch at one point inside.

§ 2. Properties and their evidence

Let us proceed directly to the proof of the properties.

§ 2.1 Property 1

The segments between the points of intersection of tangents with circles are equal to each other and equal to two geometric mean radii of these circles.

Proof 1. О 1 А 1 and О 2 В 1 - radii drawn to points of tangency.

2. О 1 А 1 ┴ А 1 В 1, О2В1 ┴ А 1 В 1 → О 1 А 1 ║ О 2 В 1. (According to point 1)

1. ▲ О 1 О 2 D - rectangular, because О 2 D ┴ О 2 В 1
2. О 1 О 2 = R + r, О 2 D = R - r

1. By the Pythagorean theorem, A 1 B 1 = 2√Rr

(O 1 D 2 = (R + r) 2 - (R-r) 2 = R 2 + 2Rr + r2-R 2 + 2Rr-r 2 = √4Rr = 2√Rr)

А 2 В 2 = 2√Rr (it is proved similarly)

1) Draw the radii at the intersection points of the tangents with the circles.

2) These radii will be perpendicular to the tangents and parallel to each other.

3) Drop the perpendicular from the center of the smaller circle to the radius of the larger circle.

4) The hypotenuse of the resulting right-angled triangle is equal to the sum of the radii of the circles. The leg is equal to their difference.

5) By the Pythagorean theorem, we obtain the required relation.

§ 2.2 Property 2

The intersection points of a straight line intersecting the tangent point of the circles and not lying in any of them with tangents divide the segments of the outer tangents, bounded by the tangency points, in half, into parts, each of which is equal to the geometric mean of the radii of these circles.

Proof 1.MC= MA 1 (as segments of tangents)

2.MC = MV 1 (as tangent line segments)

3.A 1 M = MV 1 = √Rr, A 2 N = NB 2 = √Rr (according to paragraphs 1 and 2 )

Statements used in the proof Segments of tangents drawn from one point to a certain circle are equal. Let's use this property for both given circles.

§ 2.3 Property 3

The length of the segment of the inner tangent, enclosed between the outer tangents, is equal to the length of the segment of the outer tangent between the points of tangency and is equal to two geometric mean radii of these circles.

Proof This conclusion follows from the previous property.

MN = MC + CN = 2MC = 2A 1 M = A 1 B 1 = 2√Rr

§ 2.4 Property 4

The triangle formed by the centers of the tangent circles and the midpoint of the tangent segment between the radii drawn to the tangency points is rectangular. The ratio of its legs is equal to the quotient of the roots of the radii of these circles.

Proof 1.MO 1 is the bisector of angle A 1 MC, MO 2 is the bisector of angle B 1 MC, since the center of a circle inscribed in an angle lies on the bisector of this angle.

2.According to point 1 РО 1 МС + РСМО 2 = 0.5 (РА1МС + РСМВ 1) = 0.5p = p / 2

3. PO 1 MO 2 - straight line. MS - the height of the triangle O 1 MO 2, because tangent МN is perpendicular to the radii drawn at the points of tangency → triangles О 1 МС and МО 2 С are similar.

4.O 1 M / MO 2 = O 1 S / MS = r / √Rr = √r / R (by similarity)

Statements used in the proof 1) The center of a circle inscribed in an angle lies on the bisector of this angle. The legs of the triangle are the bisectors of the angles.

2) Using the fact that the angles formed in this way are equal, we find that the desired angle we are considering is a straight line. We conclude that this triangle is indeed rectangular.

3) We prove the similarity of triangles to which the height (since the tangent is perpendicular to the radii drawn at the tangency points) divides a right-angled triangle, and in the similarity we obtain the desired ratio.

§ 2.5 Property 5

The triangle formed by the point of tangency of the circles with each other and the points of intersection of the circles with the tangent is rectangular. The ratio of its legs is equal to the quotient of the roots of the radii of these circles.

Proof

1. ▲ А 1 МС and ▲ СМВ 1 - isosceles → РМА 1 С = РМСА 1 = α, РМВ 1 С = РМСВ 1 = β.

1. 2α + 2β + РА 1 МС + РСМВ 1 = 2p → 2α + 2β = 2p - (РА 1 МС + РСМВ 1) = 2p - p = p, α + β = p / 2

1. But lA 1 CW 1 = α + β → lA 1 CW 1 - straight line → lB 1 CO 2 = lCB 1 O 2 = p / 2 - β = α

1. ▲ А 1 МС and ▲ СО 2 В 1 - similar → А 1 С / CB 1 = МС / О 2 В 1 = √Rr / R = √r / R

Statements used in the proof 1) We paint the sum of the angles of the triangles, taking advantage of the fact that they are isosceles. The isosceles of triangles is proved using the property on the equality of segments of tangents.

2) Having written the sum of the angles in this way, we find that there is a right angle in the triangle under consideration, therefore it is rectangular. The first part of the statement is proved.

3) By the similarity of triangles (when justifying it, we use the sign of similarity in two corners), we find the ratio of the legs of a right-angled triangle.

§ 2.6 Property 6

The quadrilateral formed by the points of intersection of the circles with the tangent is a trapezoid into which a circle can be inscribed.

Proof 1. ▲ А 1 RA 2 and ▲ В 1 РВ 2 - isosceles because А 1 Р = РВ 2 and В 1 Р = РВ 2 as segments of tangents → ▲ А 1 РВ 2 and ▲ В 1 РВ 2 are similar.

2.А 1 А 2 ║ В 1 В 2, because are equal to the corresponding angles formed at the intersection of the secant A 1 B 1.

1. MN - middle line by property 2 → А 1 А 2 + В 1 В 2 = 2MN = 4√Rr

1. А 1 В 1 + А 2 В 2 = 2√Rr + 2√Rr = 4√Rr = А 1 А 2 + В 1 В 2 → in the trapezoid А 2 А 1 В 1 В 2 the sum of bases is equal to the sum of the sides, and this is a necessary and sufficient condition for the existence of an inscribed circle.

Statements used in the proof 1) We again use the property of tangent line segments. With its help, we will prove the isosceles of the triangles formed by the point of intersection of tangents and points of tangency.

2) From this, the similarity of these triangles and the parallelism of their bases will follow. On this basis, we conclude that this quadrilateral is a trapezoid.

3) Using the property (2) we proved earlier, we find the middle line of the trapezoid. It is equal to two geometric mean radii of the circles. In the resulting trapezoid, the sum of the bases is equal to the sum of the sides, and this is a necessary and sufficient condition for the existence of an inscribed circle.

§ 3 Objectives

Let us consider, using a practical example, how you can simplify the solution of the problem using the above properties.

Problem 1

In the ABC triangle, the AC side = 15 cm. A circle is inscribed in the triangle. The second circle touches the first and sides AB and BC. On the AB side, point F is selected, and on the BC side - point M so that the segment FM is a common tangent to the circles. Find the ratio of the areas of triangle BFM and quadrangle AFMC, if FM is 4 cm, and point M is spaced from the center of one circle at a distance twice as large as from the center of another.

Given: FM common tangent AC = 15cm FM = 4cm O 2 M = 2O 1 M

Find S BFM / S AFMC

Solution:

1) FM = 2√Rr, ​​O 1 M / O 2 M = √r / R

2) 2√Rr = 4, √r / R = 0.5 → r = 1, R = 4; PQ = FM = 4

3) ▲ BO 1 P and ▲ BO 2 Q are similar → BP / BQ = O 1 P / O 2 Q, BP / (BP + PQ) = r / R, BP / (BP + 4) = 0.25; BP = 4/3

4) FM + BP = 16/3, S FBM = r * P FBM = 1 * (16/3) = 16/3; AC + BQ = 15 + 4/3 + 4 = 61/3

5) S ABC = R * P ABC = 4 * (61/3) = 244/3 → S BFM / S AFMC = (16/3) :( 244/3) = 4/61

Task 2

In an isosceles triangle ABC, two tangent circles are inscribed with their common point D and a common tangent FK passing through this point. Find the distance between the centers of these circles, if the base of the triangle AC = 9 cm, and the segment of the side of the triangle between the points of contact of the circles is 4 cm.

Given: ABC - isosceles triangle; FK is the common tangent of the inscribed circles. AC = 9 cm; NE = 4 cm

Solution:

Let lines AB and CD intersect at point O. Then ОА = ОD, ОВ = OC, therefore CD = = AB = 2√Rr

Points O 1 and O 2 lie on the bisector of the angle AOD. The bisector of an isosceles triangle AOD is its height, therefore AD ┴ O 1 O 2 and BC ┴ O 1 O 2, hence,

AD ║ BC and ABCD is an isosceles trapezoid.

The segment MN is its middle line, therefore AD + BC = 2MN = 2AB = AB + CD

Therefore, a circle can be inscribed in this trapezoid.

Let AP be the height of the trapezoid, right-angled triangles APB and O 1 FO 2 are similar, therefore AP / O 1 F = AB / O 1 O 2.

From this we find that

Bibliography

• Supplement to the newspaper "First September" "Mathematics" No. 43, 2003
• Unified State Exam 2010. Mathematics. Problem C4. Gordin R.K.