Ministry of Education and Science Russian Federation
Municipal budgetary educational institution
city of Novosibirsk "Gymnasium No. 4"
Section: mathematics
RESEARCH WORK
on this topic:
PROPERTIES OF TWO TOUCHING CIRCLES
10th grade students:
Khaziakhmetov Radik Ildarovich
Zubarev Evgeny Vladimirovich
Supervisor:
L.L. Barinova
Mathematic teacher
Highest qualification category
§ 1.Introduction………..………………………….……………………………………………………3
§ 1.1 Mutual arrangement two circles………………………...…………...………3
§ 2 Properties and their proofs………………………………………..…………….....….…4
§ 2.1 Property 1………………...……………………………………..…………………...….…4
§ 2.2 Property 2……………………………………………………………………………...…………………………………………………………………………………………………………………………………………………
§ 2.3 Property 3……………………………………………………..…………………...…………6
§ 2.4 Property 4…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….
§ 2.5 Property 5…………………………………..……………………………………...………8
§ 2.6 Property 6……………………………………………………………………………………………9
§ 3 Tasks…………………………………………………..…………………...……………..…11
References………………………………………………………………….………….13
§ one. Introduction
Many problems involving two tangent circles can be solved more concisely and simply by knowing some of the properties that will be presented later.
Mutual arrangement of two circles
To begin with, we will discuss the possible mutual arrangement of the two circles. There may be 4 different cases.
1. Circles may not intersect.
2. Cross.
3. Touch at one point outside.
4. Touch at one point inside.
§ 2. Properties and their proofs
Let us proceed directly to the proof of the properties.
§ 2.1 Property 1
The segments between the intersection points of the tangents with the circles are equal to each other and equal to two geometric mean radii of these circles.
Proof 1. O 1 A 1 and O 2 V 1 - radii drawn to the points of contact.
2. O 1 A 1 ┴ A 1 V 1, O2V1 ┴ A 1 V 1 → O 1 A 1 ║ O 2 V 1. (according to paragraph 1)
- ▲O 1 O 2 D - rectangular, because O 2 D ┴ O 2 V 1
- O 1 O 2 \u003d R + r, O 2 D \u003d R - r
- By the Pythagorean theorem А 1 В 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)
A 2 B 2 = 2√Rr (proved similarly)
1) Draw the radii to 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 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 desired relation.
§ 2.2 Property 2
The points of intersection of a line that intersects the point of tangency of the circles and does not lie in any of them, with tangents bisect the segments of external tangents bounded by the points of tangency, into parts, each of which is equal to the geometric mean of the radii of these circles.
Proof 1.MS= MA 1 (as segments of tangents)
2.MS = MV 1 (as segments of tangents)
3.A 1 M \u003d MV 1 \u003d √Rr, A 2 N \u003d NB 2 \u003d √Rr (according to paragraph 1 and 2 )
Statements used in the proof The segments of tangents drawn from one point to some circle are equal. We use this property for both given circles.
§ 2.3 Property 3
The length of the segment of the internal tangent enclosed between the external tangents is equal to the length of the segment of the external tangent between the points of contact 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 points of tangency 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, because The center of a circle inscribed in an angle lies on the bisector of that angle.
2. According to paragraph 1 РО 1 МS + РСМО 2 = 0.5 (РА1МС + РСМВ 1) = 0.5p = p/2
3.РО 1 MO 2 - straight. MS - the height of the triangle O 1 MO 2, because the tangent MN is perpendicular to the radii drawn to the points of contact → the triangles О 1 МС and MO 2 С are similar.
4.O 1 M / MO 2 \u003d O 1 C / MS \u003d r / √Rr \u003d √r / R (by similarity)
Statements used in the proof 1) The center of a circle inscribed in an angle lies on the bisector of that angle. The legs of a triangle are the bisectors of the angles.
2) Using the fact that the angles formed in this way are equal, we obtain that the angle we are looking for is a right angle. We conclude that this triangle is indeed a right triangle.
3) We prove the similarity of the triangles into which the height (since the tangent is perpendicular to the radii drawn at the points of contact) divides the right triangle, and by similarity we obtain the desired ratio.
§ 2.5 Property 5
The triangle formed by the point of contact of the circles with each other and the points of intersection of the circles with the tangent, is a right triangle. The ratio of its legs is equal to the quotient of the roots of the radii of these circles.
Proof
- ▲А 1 МС and ▲СМВ 1 are isosceles → РМА 1 С = РМСА 1 = α, РМВ 1 С = РМСВ 1 = β.
- 2α + 2β + РА 1 MS + РСМВ 1 = 2p → 2α + 2β = 2p - (РА 1 МS + РСМВ 1) = 2p - p = p, α + β = p/2
- But RA 1 SV 1 = α + β → RA 1 SV 1 - direct → РВ 1 CO 2 = РSV 1 О 2 = p/2 - β = α
- ▲A 1 MS and ▲CO 2 B 1 are similar → A 1 C / SV 1 = MS / O 2 B 1 = √Rr / R = √r / R
Statements used in the proof 1) We paint the sum of the angles of triangles, using the fact that they are isosceles. The isosceles triangles are proved using the property about the equality of tangent segments.
2) Having painted the sum of the angles in such a way, we get that in the triangle under consideration there is a right angle, 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 at two angles) we find the ratio of the legs of a right triangle.
§ 2.6 Property 6
The quadrilateral formed by the intersection points of the circles with the tangent is a trapezoid into which the circle can be inscribed.
Proof 1.▲A 1 RA 2 and ▲B 1 RV 2 are isosceles because A 1 P \u003d RA 2 and B 1 P \u003d PB 2 as segments of tangents → ▲A 1 RA 2 and ▲B 1 PB 2 are similar.
2.A 1 A 2 ║ B 1 B 2, because the corresponding angles formed at the intersection of the secant A 1 B 1 are equal.
- MN - middle line by property 2 → A 1 A 2 + B 1 B 2 = 2MN = 4√Rr
- A 1 B 1 + A 2 B 2 \u003d 2 √ Rr + 2 √ Rr \u003d 4 √ Rr \u003d A 1 A 2 + B 1 B 2 → in a trapezoid A 2 A 1 B 1 B 2 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.
Statements used in the proof 1) Let's use the property of tangent segments again. With its help, we will prove the isosceles triangles formed by the intersection point of the tangents and the tangent points.
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) According to the property (2) we proved earlier, we find the median line of the trapezoid. It is equal to two geometric mean radii of 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. Tasks
Consider, using a practical example, how the solution of the problem can be simplified using the above properties.
Task 1
In triangle ABC, side AC = 15 cm. A circle is inscribed in the triangle. The second circle touches the first and the sides AB and BC. Point F is chosen on side AB, and point M is chosen on side BC so that segment FM is a common tangent to the circles. Find the ratio of the areas of triangle BFM and quadrilateral AFMC if FM is 4 cm, and point M is twice as far from the center of one circle as it is from the center of the other.
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 \u003d R * R ABC \u003d 4 * (61/3) \u003d 244/3 → S BFM / S AFMC \u003d (16/3): (244/3) \u003d 4/61
Task 2
Two tangent circles with their common point D and a common tangent FK passing through this point are inscribed in an isosceles triangle ABC. Find the distance between the centers of these circles if the base of the triangle AC = 9 cm, and the segment of the lateral side of the triangle enclosed between the points of contact of the circles is 4 cm.
Given: ABC is an 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 OA = OD, OB = OC, so CD = AB = 2√Rr
Points O 1 and O 2 lie on the bisector of the angle AOD. Bisector isosceles triangle AOD is its height, so AD ┴ O 1 O 2 and BC ┴ O 1 O 2, so
AD ║ BC and ABCD is an isosceles trapezoid.
The segment MN is its midline, so AD + BC = 2MN = 2AB = AB + CD
Therefore, a circle can be inscribed in this trapezoid.
Let AP be the height of the trapezoid, right triangles АРВ and О 1 FO 2 are similar, therefore АР/О 1 F = АВ/О 1 О 2 .
From here we find that 
Bibliography
- Supplement to the newspaper "First of September" "Mathematics" No. 43, 2003
- USE 2010. Mathematics. Task C4. Gordin R.K.
Lesson topic: " Mutual arrangement of two circles on a plane.
Target :
educational - mastering new knowledge about the relative position of two circles, preparing for control work
Educational - development of computational skills, development of logical and structural thinking; formation of skills for finding rational solutions and achieving final results; development cognitive activity and creative thinking.
Educational – the formation of students' responsibility, consistency; development of cognitive and aesthetic qualities; formation of information culture of students.
Correctional - develop spatial thinking, memory, hand motor skills.
Lesson type: learning new educational material, fastening.
Type of lesson: 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 of basic knowledge.
What topics did we cover in the previous lessons?
General view of the circle equation?
Perform orally:
Blitz Poll
3. Introduction of new material.
What do you think and what figure we will consider today .... What if there are two?
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 the two circles. And find out what is the distance between the centers depending on the location.
Lesson topic:« Mutual arrangement of two circles. Problem solving.»
1. Concentric circles
2. Non-intersecting circles
3.External touch
4. Intersecting circles
5. Internal touch
So let's conclude
4. Formation of skills and abilities
Find an error in the data or in the statement and correct it by giving reasons for your opinion:
a) Two circles are touching. Their radii are R = 8 cm and r = 2 cm, the distance between the centers is d = 6.
B) Two circles have at least two points in common.
C) R = 4, r = 3, d = 5. The circles have no common points.
D) R \u003d 8, r \u003d 6, d \u003d 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 touch 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 touch internally. 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 touch internally. 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.
CHECKING UNDERSTANDING
1) How can two circles be located?
2) When 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 there is time)
1)E(4;12), F(-4;-10), G(-2;6), H(4;-2) Find:
a) coordinates of the vectors EF,GH
b) the length of the vector FG
c) coordinates of the point O - the middle of EF
coordinates of point W - midpoint GH
d) circle equation with diameter FG
e) equation of the straight line FH
6. Homework
& 96 #1000. Which of these equations are circle equations. Find Center and Radius
7. Summing up the lesson(3 min.)
(give a qualitative assessment of the work of the class and individual students).
8. Stage of reflection(2 minutes.)
(initiate students' reflection on their own emotional state, their activities, interaction with the teacher and classmates using drawings)
Let a circle and a point not coinciding with its center C be given (Fig. 205). Three cases are possible: the point lies inside the circle (Fig. 205, a), on the circle (Fig. 205, b), outside the circle (Fig. 205, c). Let's draw a straight line, it will intersect the circle at points K and L (in case b), the point will coincide with one of which will be the closest to the point in comparison with all other points of the circle), and the other - the most remote.
So, for example, in Fig. 205, and the point K of the circle is the closest to . Indeed, for any other point on the circle, the broken line is longer than the segment CAG: but also, therefore, on the contrary, for the point L we find (again, the broken line is longer than the straight line segment). We leave the analysis of the remaining two cases to the reader. Note that the largest distance is equal to the smallest if or if .
Let's pass to the analysis of possible cases of an arrangement of two circles (fig. 206).
a) The centers of the circles coincide (Fig. 206, a). Such circles are called concentric. If the radii of these circles are not equal, then one of them lies inside the other. If the radii are equal, they coincide.
b) Now let the centers of the circles be different. We connect them with a straight line, it is called the line of centers of a given pair of circles. The mutual arrangement of the circles will depend only on the ratio between the value of the segment d connecting their centers and the values of the radii of the circles R, r. All possible essentially different cases are shown in fig. 206 (we consider).
1. The distance between the centers is less than the difference in radii:
(Fig. 206, b), a small circle lies inside a large one. This also includes the case a) coincidence of centers (d = 0).
2. The distance between the centers is equal to the difference of the radii:
(Fig. 206, s). The small circle lies inside the large one, but has one common point with it on the line of centers (they say that there is internal contact).
3. The distance between the centers is greater than the difference of the radii, but less than their sum:
(Fig. 206, d). Each of the circles lies partly inside, partly outside the other.
Circles have two intersection points K and L, located symmetrically about the line of centers. A segment is a common chord of two intersecting circles. It is perpendicular to the line of centers.
4. The distance between the centers is equal to the sum of the radii:
(Fig. 206, e). Each of the circles lies outside the other, but they have a common point on the line of centers (external tangency).
5. The distance between the centers is greater than the sum of the radii: (Fig. 206, e). Each of the circles lies entirely outside the other. Circles do not have common points.
The above classification follows completely from the above. above the question about the largest and smallest distance from a point to a circle. It is only necessary to consider two points on one of the circles: the closest and the farthest from the center of the second circle. For example, consider the case By condition . But the point of the small circle farthest from O is at a distance from the center O. Therefore, the entire small circle lies inside the large one. Other cases are considered in the same way.
In particular, if the radii of the circles are equal, then only the last three cases are possible: intersection, external touch, external location.
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 towards the center) a and b. Moreover, Oa and Ob are their centers. Without loss of generality, we will assume that Ra > Rb.
Then the following conditions are satisfied:
Task 1: Mansions of important noblesPoints of intersection of two circles
Suppose A and B intersect at two points. Let's find these intersection points.
To do this, the vector from a to the 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 codirectional 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.
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 rotating from p. Because we know all the sides of the triangles OaI1Ob and OaI2Ob (Radius and distance between centers), we can get this angle fi, turning the vector p in one direction will give I1, and in the other I2.
According to the law of cosines, it is equal to:
If you rotate p by fi, then you get i1 or i2, depending on which way to turn. Next, 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 lies inside the other. But there, exactly, the vector p will have to be set in the direction from a to b, which is what we did. If you build p based on another circle, then nothing will come of it
Well, in conclusion, one fact must be mentioned to everything: if the circles touch, then it is easy to make sure that P is the point of contact (this is true for both internal and external touch).
Here you can see the visualization (click to run).
Task 2: Intersection points
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 rotating the vector. This will greatly simplify the calculations, saving the code from trigonometric functions.
