Constructing a tangent to tangent circles. The relative position of two circles

Usually in such a problem you are given a circle and a point. It is required to construct a tangent to a circle, and the tangent must pass through a given point.

If the location of the point is not specified, then three possible cases of location of the point should be separately specified.

If a point lies inside a circle bounded by a given circle, then a tangent cannot be constructed through it.
If a point lies on a circle, then the tangent is constructed by constructing a line perpendicular to the radius drawn to the given point.
If a point lies outside the circle bounded by a circle, then before constructing a tangent, a point on the circle is sought through which it must pass.

To solve the second case, on the straight line on which the radius lies, a segment is constructed that is equal to the radius and lies on the other side of the point on the circle. Thus, a point on a circle turns out to be the middle of a segment equal to twice the radius. Next, two circles are constructed whose radii are equal to twice the radius of the original circle, with centers at the ends of the segment equal to twice the radius. A straight line is drawn through any point of intersection of these circles and a point specified by the conditions of the problem. It will be the median perpendicular to the radius of the original circle, that is, perpendicular to it, and therefore tangent to the circle.

You can solve the third case, when the point lies outside the circle bounded by the circle, as follows. It is necessary to construct a segment connecting the center of a given circle and a given point. Next, find its middle by constructing a median perpendicular (described in the previous paragraph). After this, draw a circle (or part of it). The intersection point of the constructed circle and the one specified by the problem conditions is the point through which the tangent passes, which also passes through the point specified by the problem conditions. A tangent line is drawn through two known points.

To prove that the constructed straight line is a tangent, one should consider the angle formed by the radius of the circle given by the conditions of the problem and the segment connecting the point of intersection of the circles with the point given by the conditions of the problem. This angle rests on a semicircle (the diameter of the constructed circle), which means it is straight. That is, the radius is perpendicular to the constructed line. Therefore, the constructed line is tangent.

Lesson Objectives

Educational – repetition, generalization and testing of knowledge on the topic: “Tangent to a circle”; development of basic skills.
Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.
Introduce the concept of a tangent, a point of contact.
Consider the property of a tangent and its sign and show their application in solving problems in nature and technology.

Lesson Objectives

Develop skills in constructing tangents using a scale ruler, protractor and drawing triangle.
Test students' problem-solving skills.
Ensure mastery of the basic algorithmic techniques for constructing a tangent to a circle.
Develop the ability to apply theoretical knowledge to problem solving.
Develop students' thinking and speech.
Work on developing the skills to observe, notice patterns, generalize, and reason by analogy.
Instilling an interest in mathematics.

Lesson Plan

The emergence of the concept of tangent.
The history of the tangent.
Geometric definitions.
Basic theorems.
Constructing a tangent to a circle.
Consolidation.

The emergence of the concept of tangent

The concept of a tangent is one of the oldest in mathematics. In geometry, a tangent to a circle is defined as a line that has exactly one point of intersection with that circle. The ancients, using compasses and rulers, were able to draw tangents to a circle, and later to conic sections: ellipses, hyperbolas and parabolas.

The history of the tangent

Interest in tangents was revived in modern times. Then curves were discovered that were unknown to ancient scientists. For example, Galileo introduced the cycloid, and Descartes and Fermat constructed a tangent to it. In the first third of the 17th century. They began to understand that a tangent is a straight line, “most closely adjacent” to a curve in a small neighborhood of a given point. It is easy to imagine a situation where it is impossible to construct a tangent to the curve at a given point (figure).

Geometric definitions

Circle- the geometric locus of points on the plane equidistant from a given point, called its center.

circle.

Related definitions

A segment connecting the center of a circle with any point on it (as well as the length of this segment) is called radius circles.
The part of the plane bounded by a circle is called all around.
A segment connecting two points on a circle is called its chord. A chord passing through the center of a circle is called diameter.
Any two divergent points on a circle divide it into two parts. Each of these parts is called arc circles. The measure of an arc can be the measure of its corresponding central angle. An arc is called a semicircle if the segment connecting its ends is a diameter.
A straight line that has exactly one common point with a circle is called tangent to a circle, and their common point is called the tangency point of the line and the circle.
A straight line passing through two points on a circle is called secant.
A central angle in a circle is a plane angle with a vertex at its center.
An angle whose vertex lies on a circle and whose sides intersect this circle is called inscribed angle.
Two circles having a common center are called concentric.

Tangent line- a straight line passing through a point on a curve and coinciding with it at this point up to first order.

Tangent to a circle is a straight line that has one common point with a circle.

A straight line passing through a point on a circle in the same plane perpendicular to the radius drawn to this point called tangent. In this case, this point on the circle is called the point of tangency.

Where in our cases “a” is a straight line which is tangent to a given circle, point “A” is the point of tangency. In this case, a⊥OA (straight line a is perpendicular to the radius OA).

They say that two circles touch, if they have a single common point. This point is called point of contact of the circles. Through the point of contact, you can draw a tangent to one of the circles, which is also a tangent to the other circle. Touching circles can be internal or external.

A tangency is called internal if the centers of the circles lie on the same side of the tangent.

A tangency is called external if the centers of the circles lie on opposite sides of the tangent

a is the common tangent to the two circles, K is the point of tangency.

Basic theorems

Theorem about tangent and secant

If a tangent and a secant are drawn from a point lying outside the circle, then the square of the length of the tangent is equal to the product of the secant and its outer part: MC 2 = MA MB.

Theorem. The radius drawn to the point of tangency of the circle is perpendicular to the tangent.

Theorem. If the radius is perpendicular to a line at the point where it intersects a circle, then this line is tangent to this circle.

Proof.

To prove these theorems, we need to remember what a perpendicular from a point to a line is. This is the shortest distance from this point to this line. Let us assume that OA is not perpendicular to the tangent, but there is a straight line OS perpendicular to the tangent. The length OS includes the length of the radius and a certain segment BC, which is certainly greater than the radius. Thus, one can prove it for any line. We conclude that the radius, the radius drawn to the point of contact, is the shortest distance to the tangent from point O, i.e. OS is perpendicular to the tangent. In the proof of the converse theorem, we will proceed from the fact that the tangent has only one common point with the circle. Let this straight line have one more common point B with the circle. Triangle AOB is rectangular and its two sides are equal as radii, which cannot be the case. Thus, we find that this straight line has no more points in common with the circle except point A, i.e. is tangent.

Theorem. The tangent segments drawn from one point to the circle are equal, and the straight line connecting this point with the center of the circle divides the angle between the tangents.

Proof.

The proof is very simple. Using the previous theorem, we assert that OB is perpendicular to AB, and OS is perpendicular to AC. Right triangles ABO and ACO are equal in leg and hypotenuse (OB=OS - radii, AO - total). Therefore, their sides AB=AC and angles OAC and OAB are equal.

Theorem. The magnitude of the angle formed by a tangent and a chord having a common point on a circle is equal to half the angular magnitude of the arc enclosed between its sides.

Proof.

Consider the angle NAB formed by a tangent and a chord. Let's draw the diameter of AC. The tangent is perpendicular to the diameter drawn to the point of contact, therefore, ∠CAN=90 o. Knowing the theorem, we see that angle alpha (a) is equal to half the angular value of the arc BC or half the angle BOS. ∠NAB=90 o -a, from here we get ∠NAB=1/2(180 o -∠BOC)=1/2∠AOB or = half the angular value of the arc BA. etc.

Theorem. If a tangent and a secant are drawn from a point to a circle, then the square of the tangent segment from a given point to the point of tangency is equal to the product of the lengths of the secant segments from a given point to the points of its intersection with the circle.

Proof.

In the figure, this theorem looks like this: MA 2 = MV * MC. Let's prove it. According to the previous theorem, the angle MAC is equal to half the angular value of the arc AC, but also the angle ABC is equal to half the angular value of the arc AC according to the theorem, therefore, these angles are equal to each other. Taking into account the fact that triangles AMC and BMA have a common angle at the vertex M, we state the similarity of these triangles in two angles (second sign). From the similarity we have: MA/MB=MC/MA, from which we get MA 2 =MB*MC

Constructing tangents to a circle

Now let's try to figure it out and find out what needs to be done to construct a tangent to a circle.

In this case, as a rule, the problem gives a circle and a point. And you and I need to construct a tangent to the circle so that this tangent passes through a given point.

In the event that we do not know the location of a point, then let's consider cases of possible locations of points.

Firstly, a point may be inside a circle, which is limited by a given circle. In this case, it is not possible to construct a tangent through this circle.

In the second case, the point is located on a circle, and we can construct a tangent by drawing a perpendicular line to the radius, which is drawn to the point known to us.

Thirdly, let’s assume that the point is located outside the circle, which is limited by the circle. In this case, before constructing a tangent, it is necessary to find a point on the circle through which the tangent must pass.

With the first case, I hope everything is clear to you, but to solve the second option we need to construct a segment on the straight line on which the radius lies. This segment must be equal to the radius and the segment that lies on the circle on the opposite side.

Here we see that a point on a circle is the middle of a segment that is equal to twice the radius. The next step will be to construct two circles. The radii of these circles will be equal to twice the radius of the original circle, with centers at the ends of the segment, which is equal to twice the radius. Now we can draw a straight line through any point of intersection of these circles and a given point. Such a straight line is the median perpendicular to the radius of the circle that was drawn initially. Thus, we see that this line is perpendicular to the circle and it follows from this that it is tangent to the circle.

In the third option, we have a point lying outside the circle, which is limited by a circle. In this case, we first construct a segment that will connect the center of the provided circle and the given point. And then we find its middle. But for this it is necessary to construct a perpendicular bisector. And you already know how to build it. Then we need to draw a circle, or at least part of it. Now we see that the point of intersection of the given circle and the newly constructed one is the point through which the tangent passes. It also passes through the point that was specified according to the conditions of the problem. And finally, through the two points you know, you can draw a tangent line.

And finally, in order to prove that the straight line we constructed is a tangent, we need to pay attention to the angle that was formed by the radius of the circle and the segment known by the condition and connecting the point of intersection of the circles with the point given by the condition of the problem. Now we see that the resulting angle rests on a semicircle. And from this it follows that this angle is right. Consequently, the radius will be perpendicular to the newly constructed line, and this line is the tangent.

Construction of a tangent.

The construction of tangent lines is one of those problems that led to the birth of differential calculus. The first published work related to differential calculus, written by Leibniz, was entitled “A new method of maxima and minima, as well as tangents, for which neither fractional nor irrational quantities, nor a special type of calculus, are an obstacle.”

Geometric knowledge of the ancient Egyptians.

If we do not take into account the very modest contribution of the ancient inhabitants of the valley between the Tigris and Euphrates and Asia Minor, then geometry originated in Ancient Egypt before 1700 BC. During the tropical rainy season, the Nile replenished its water reserves and overflowed. Water covered areas of cultivated land, and for tax purposes it was necessary to determine how much land was lost. Surveyors used a tightly stretched rope as a measuring tool. Another incentive for the accumulation of geometric knowledge by the Egyptians was their activities such as the construction of pyramids and fine arts.

The level of geometric knowledge can be judged from ancient manuscripts, which are specifically devoted to mathematics and are something like textbooks, or rather, problem books, where solutions to various practical problems are given.

The oldest mathematical manuscript of the Egyptians was copied by a certain student between 1800 - 1600. BC. from an older text. The papyrus was found by the Russian Egyptologist Vladimir Semenovich Golenishchev. It is kept in Moscow - in the Museum of Fine Arts named after A.S. Pushkin, and is called the Moscow papyrus.

Another mathematical papyrus, written two to three hundred years later than Moscow’s, is kept in London. It is called: “Instruction on how to achieve knowledge of all dark things, all the secrets that things hide in themselves... According to old monuments, the scribe Ahmes wrote this.” The manuscript is called the “Ahmes papyrus”, or the Rhind papyrus - after the name of the Englishman who found and bought this papyrus in Egypt. The Ahmes papyrus provides solutions to 84 problems involving various calculations that may be needed in practice.

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Ministry of Education and Science of the Russian Federation

Municipal budgetary educational institution

city of Novosibirsk "Gymnasium No. 4"

Section: mathematics

RESEARCH

on this topic:

PROPERTIES OF TWO TOUCHING CIRCLES

10th grade students:

Khaziakhmetov Radik Ildarovich

Zubarev Evgeniy Vladimirovich

Supervisor:

L.L. Barinova

Mathematic teacher

Highest qualification category

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

§ 1.1 The relative position of two circles………………………...………………...………3

§ 2 Properties and their evidence…………………………………………………………..………………….....….…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 more briefly and simply by knowing some of the properties that will be presented next.

The relative position of two circles

To begin with, let us stipulate the possible relative position of the two circles. There may be 4 different cases.

1. The circles may not intersect.

2. Intersect.

3. Touch at one point on the outside.

4.Touch at one point inside.

§ 2. Properties and their proofs

Let's move directly to the proof of the properties.

§ 2.1 Property 1

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

Proof 1. O 1 A 1 and O 2 B 1 – radii drawn to the points of contact.

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)

▲O 1 O 2 D – rectangular, because О 2 D ┴ О 2 В 1
O 1 O 2 = R + r, O 2 D = R – r

According to 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)

A 2 B 2 = 2√Rr (proved similarly)

1) Let's draw the radii at the points of intersection of the tangents with the circles.

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

3) Let us lower a 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) Using the Pythagorean theorem we obtain the required relationship.

§ 2.2 Property 2

The points of intersection of a straight line that intersects the tangent point of the circles and does not lie in any of them with the tangents divide in half the segments of the external tangents, limited 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 tangent segments)

2.MC = MV 1 (as tangent segments)

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

Statements used in the proof The tangent segments drawn from one point to a certain 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 the given 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 tangent circles and the midpoint of the tangent segment between the radii drawn to the points of contact 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 MS, MO 2 is the bisector of angle B 1 MS, because The center of a circle inscribed in an angle lies on the bisector of this angle.

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

3.РО 1 MO 2 – direct. MC is the height of the triangle O 1 MO 2, because the tangent MN is perpendicular to the radii drawn to the points of contact → triangles O 1 MC and MO 2 C are similar.

4.O 1 M / MO 2 = O 1 C / MC = r / √Rr = √r / R (similar)

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 a triangle are the bisectors of the angles.

2) Using the fact that the angles formed in this way are equal, we find that the angle we are looking for is a right angle. We conclude that this triangle is indeed right-angled.

3) We prove the similarity of the triangles into which the height (since the tangent is perpendicular to the radii drawn to the points of tangency) divides the right triangle, and by similarity we obtain the required 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 rectangular. The ratio of its legs is equal to the quotient of the roots of the radii of these circles.

Proof

▲A 1 MC and ▲SMV 1 are isosceles → ÐMA 1 C = ÐMSA 1 = α, ÐMV 1 C = ÐMSV 1 = β.

2α + 2β + RA 1 MC + RSMV 1 = 2p → 2α + 2β = 2p - (RA 1 MC + RSMV 1) = 2p - p = p, α + β = p/2

But RA 1 SV 1 = α + β → RA 1 SV 1 – direct → RA 1 CO 2 = RS 1 O 2 = p/2 – β = α

▲A 1 MC and ▲CO 2 B 1 are similar → A 1 C / SV 1 = MC / O 2 B 1 = √Rr / R = √r / R

Statements used in the proof 1) We write down 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 of equality of tangent segments.

2) Having written the sum of angles in this way, we find that the triangle in question has a right angle, therefore it is rectangular. The first part of the statement has been proven.

3) Using the similarity of triangles (to justify 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 points of intersection of the circles with the tangent is a trapezoid into which a circle can be inscribed.

Proof 1.▲A 1 RA 2 and ▲B 1 PB 2 are isosceles because A 1 P = RA 2 and B 1 P = PB 2 as tangent segments → ▲A 1 RA 2 and ▲B 1 PB 2 – 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 according to property 2 → A 1 A 2 + B 1 B 2 = 2MN = 4√Rr

A 1 B 1 + A 2 B 2 = 2√Rr + 2√Rr = 4√Rr = A 1 A 2 + B 1 B 2 → in the 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 us again use the property of tangent segments. With its help, we will prove the isosceles of triangles formed by the intersection point of tangents and points of tangency.

2) From this it will follow that these triangles are similar and their bases are parallel. On this basis we conclude that this quadrilateral is a trapezoid.

3) Using the property (2) we proved earlier, we find the midline 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. Problems

Let's look at a practical example of how you can simplify the solution of a problem using the properties outlined above.

Problem 1

In triangle ABC, side AC = 15 cm. A circle is inscribed in the triangle. The second circle touches the first and sides AB and BC. On side AB, point F is selected, and on side BC, point M is selected 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 located twice as far from the center of one circle as from the center of the other.

Given: FM-total tangent AC=15cm FM=4cm O 2 M=2О 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

Problem 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 side of the triangle enclosed between the points of tangency of the circles is 4 cm.

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

Solution:

Let straight 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 angle AOD. The bisector of an isosceles triangle AOD is its altitude, so AD ┴ O 1 O 2 and BC ┴ O 1 O 2, which means

AD ║ BC and ABCD – isosceles trapezoid.

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 ARB and O 1 FO 2 are similar, therefore AP/O 1 F = AB/O 1 O 2 .

From here we find that

Bibliography

Supplement to the newspaper “First of September” “Mathematics” No. 43, 2003
Unified State Exam 2010. Mathematics. Task C4. Gordin R.K.