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Construction of a direct tangent to two circles. Tangent to a circle. Full lessons - Knowledge Hypermarket. Highest qualification category

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 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……………………………………………………………………………...…………………………………………………………………………………………………………………………………………………

§ 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)



  1. ▲O 1 O 2 D - rectangular, because O 2 D ┴ O 2 V 1
  2. O 1 O 2 \u003d R + r, O 2 D \u003d R - r

  1. 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

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

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

  1. But RA 1 SV 1 = α + β → RA 1 SV 1 - direct → РВ 1 CO 2 = РSV 1 О 2 = p/2 - β = α

  1. ▲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.

  1. MN - middle line by property 2 → A 1 A 2 + B 1 B 2 = 2MN = 4√Rr

  1. 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

Decision:

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

Decision:

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. The bisector of an 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.

Geometric constructions

Construction of tangents to circles

Consider the problem underlying the solution of other problems on drawing tangents to circles.

Let from the pointBUT(Fig. 1) it is necessary to draw tangents to a circle centered at a pointO.

To accurately construct tangents, it is necessary to determine the points of tangency of the lines to the circle. For this pointBUTshould be connected with a dotOand split the segmentOAin half. From the middle of this segment - pointsWith, how to describe a circle from the center, the diameter of which should be equal to the segmentOA. pointsTo1 andTo2 intersections of circles centered at a pointWithand centered at a pointOare the points of contact of the linesAK1 andAK2 to a given circle.

The correctness of the solution of the problem is confirmed by the fact that the radius of the circle drawn to the point of contact is perpendicular to the tangent to the circle. cornersOK1 BUTandOK2 BUTare straight because they rely on the diameterJSCcircle centered at a pointWith.

Rice. one.

When constructing tangents to two circles, tangents are distinguisheddomesticandexternal. If the centers of the given circles are located on one side of the tangent, then it is considered external, and if the centers of the circles are on opposite sides of the tangent, it is considered internal.

O1 andO2 R1 andR2 . It is required to draw external tangents to given circles.

For precise construction, it is necessary to determine the points of contact between the lines and given circles. If the radii of circles with centersO1 andO2 start successively decreasing by the same value, then you can get a series of concentric circles of smaller diameters. Moreover, in each case of a decrease in the radius, the tangents to the smaller circles will be parallel to the desired ones. After reducing both radii by the size of the smaller radiusR2 circle with centerO2 will turn into a point, and a circle with a centerO1 will be transformed into a concentric circle with a radiusR3 , equal to the difference of radiiR1 andR2 .

Using the method described earlier, from the pointO2 draw external tangents to a circle with radiusR3 , connect the dotsO1 andO2 , divided by a dotWithline segmentO1 O2 in half and draw a radiusSO1 an arc whose intersection with a given circle will determine the points of contact of the linesO2 To1 andO2 To2 .

DotBUT1 andBUT2 the contact of the desired lines with a larger circle is located on the continuation of the linesO1 To1 andO1 To2 . pointsAT1 andAT2 tangents of lines with a smaller circle are perpendicular to the baseO2 respectively to the auxiliary tangentsO2 To1 andO2 To2 . Having points of contact, you can draw the desired linesBUT1 AT1 andBUT2 AT2 .

Rice. 2.

Let two circles with centers at the pointsO1 andO2 (Fig. 2), having radii, respectivelyR1 andR2 . It is required to draw internal tangents to given circles.

To determine the points of contact between lines and circles, we use arguments similar to those given in solving the previous problem. If we decrease the radiusR2 to zero, then the circle with the centerO2 turn to the point. However, in this case, in order to preserve the parallelism of the auxiliary tangents with the required ones, the radiusR1 should be enlargedR2 and draw a circle with a radiusR3 , equal to the sum of the radiiR1 andR2 .

From a pointO2 draw tangents to a circle with a radiusR3 , for which we connect the dotsO1 andO2 , divided by a dotWithline segmentO1 O2 in half and draw an arc of a circle centered at a pointWithand radiusSO1 . Intersection of an arc with a circle of radiusR3 will determine the position of the pointsTo1 andTo2 tangency of auxiliary linesO2 To1 andO2 To2 .

DotBUT1 andBUT2 R1 is at the intersection of this circle with the segmentO1 To1 andO1 To2 . To define pointsIN 1andIN 2tangency of the desired lines with a circle of radiusR2 follows from the pointO2set up perpendiculars to auxiliary linesO2K1andO2K2until it intersects with a given circle. Having the tangent points of the desired lines and given circles, we draw the linesA1B1andA2B2.

Rice. 3.

Transects, tangents - all this could be heard hundreds of times in geometry lessons. But graduation from school is over, years pass, and all this knowledge is forgotten. What should be remembered?

Essence

The term "tangent to a circle" is probably familiar to everyone. But it is unlikely that everyone will be able to quickly formulate its definition. Meanwhile, a tangent is such a straight line lying in the same plane with a circle that intersects it at only one point. There may be a huge variety of them, but they all have the same properties, which will be discussed below. As you might guess, the point of contact is the place where the circle and the line intersect. In each case, it is one, but if there are more of them, then it will be a secant.

History of discovery and study

The concept of tangent appeared in antiquity. The construction of these straight lines, first to a circle, and then to ellipses, parabolas and hyperbolas with the help of a ruler and a compass, was carried out even at the initial stages of the development of geometry. Of course, history has not preserved the name of the discoverer, but it is obvious that even at that time people were quite aware of the properties of a tangent to a circle.

In modern times, interest in this phenomenon flared up again - a new round of studying this concept began, combined with the discovery of new curves. So, Galileo introduced the concept of a cycloid, and Fermat and Descartes built a tangent to it. As for circles, it seems that there are no secrets left for the ancients in this area.

Properties

The radius drawn to the point of intersection will be

the main, but not the only property that a tangent to a circle has. Another important feature includes already two straight lines. So, through one point lying outside the circle, two tangents can be drawn, while their segments will be equal. There is another theorem on this topic, but it is rarely covered in the framework of a standard school course, although it is extremely convenient for solving some problems. It sounds like this. From one point located outside the circle, a tangent and a secant are drawn to it. Segments AB, AC and AD are formed. A is the intersection of lines, B is the point of contact, C and D are the intersections. In this case, the following equality will be valid: the length of the tangent to the circle, squared, will be equal to the product of segments AC and AD.

There is an important consequence of the above. For each point of the circle, you can build a tangent, but only one. The proof of this is quite simple: theoretically dropping a perpendicular from the radius onto it, we find out that the formed triangle cannot exist. And this means that the tangent is unique.

Building

Among other tasks in geometry, there is a special category, as a rule, not

favored by pupils and students. To solve tasks from this category, you only need a compass and a ruler. These are building tasks. There are also methods for constructing a tangent.

So, given a circle and a point lying outside its boundaries. And it is necessary to draw a tangent through them. How to do it? First of all, you need to draw a segment between the center of the circle O and a given point. Then, using a compass, divide it in half. To do this, you need to set the radius - a little more than half the distance between the center of the original circle and the given point. After that, you need to build two intersecting arcs. Moreover, the radius of the compass does not need to be changed, and the center of each part of the circle will be the initial point and O, respectively. The intersections of the arcs must be connected, which will divide the segment in half. Set a radius on the compass equal to this distance. Next, with the center at the intersection point, draw another circle. Both the initial point and O will lie on it. In this case, there will be two more intersections with the circle given in the problem. They will be the touch points for the initially given point.

It was the construction of tangents to the circle that led to the birth

differential calculus. The first work on this topic was published by the famous German mathematician Leibniz. He provided for the possibility of finding maxima, minima and tangents, regardless of fractional and irrational values. Well, now it is used for many other calculations as well.

In addition, the tangent to the circle is related to the geometric meaning of the tangent. That is where its name comes from. Translated from Latin, tangens means "tangent". Thus, this concept is connected not only with geometry and differential calculus, but also with trigonometry.

Two circles

A tangent does not always affect only one figure. If a huge number of straight lines can be drawn to one circle, then why not vice versa? Can. But the task in this case is seriously complicated, because the tangent to two circles may not pass through any points, and the relative position of all these figures can be very

different.

Types and varieties

When it comes to two circles and one or more straight lines, even if it is known that these are tangents, it does not immediately become clear how all these figures are located in relation to each other. Based on this, there are several varieties. So, circles can have one or two common points or not have them at all. In the first case, they will intersect, and in the second, they will touch. And here there are two varieties. If one circle is, as it were, embedded in the second, then the touch is called internal, if not, then external. You can understand the relative position of the figures not only based on the drawing, but also having information about the sum of their radii and the distance between their centers. If these two quantities are equal, then the circles touch. If the first is greater, they intersect, and if less, then they do not have common points.

Same with straight lines. For any two circles that do not have common points, one can

build four tangents. Two of them will intersect between the figures, they are called internal. A couple of others are external.

If we are talking about circles that have one common point, then the task is greatly simplified. The fact is that for any mutual arrangement in this case, they will have only one tangent. And it will pass through the point of their intersection. So the construction of the difficulty will not cause.

If the figures have two points of intersection, then a straight line can be constructed for them, tangent to the circle, both one and the second, but only the outer one. The solution to this problem is similar to what will be discussed below.

Problem solving

Both internal and external tangents to two circles are not so simple in construction, although this problem can be solved. The fact is that an auxiliary figure is used for this, so think of this method yourself

quite problematic. So, given two circles with different radii and centers O1 and O2. For them, you need to build two pairs of tangents.

First of all, near the center of the larger circle, you need to build an auxiliary one. In this case, the difference between the radii of the two initial figures must be established on the compass. Tangents to the auxiliary circle are built from the center of the smaller circle. After that, from O1 and O2, perpendiculars are drawn to these lines until they intersect with the original figures. As follows from the main property of the tangent, the desired points on both circles are found. The problem is solved, at least, its first part.

In order to construct the internal tangents, one has to solve practically

a similar task. Again, an auxiliary figure is needed, but this time its radius will be equal to the sum of the original ones. Tangents are constructed to it from the center of one of the given circles. The further course of the solution can be understood from the previous example.

Tangent to a circle or even two or more is not such a difficult task. Of course, mathematicians have long ceased to solve such problems manually and trust the calculations to special programs. But do not think that now it is not necessary to be able to do it yourself, because in order to correctly formulate a task for a computer, you need to do and understand a lot. Unfortunately, there are fears that after the final transition to the test form of knowledge control, construction tasks will cause more and more difficulties for students.

As for finding common tangents for more circles, this is not always possible, even if they lie in the same plane. But in some cases it is possible to find such a line.

Real life examples

A common tangent to two circles is often encountered in practice, although this is not always noticeable. Conveyors, block systems, pulley transmission belts, thread tension in a sewing machine, and even just a bicycle chain - all these are examples from life. So do not think that geometric problems remain only in theory: in engineering, physics, construction and many other areas, they find practical application.

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State budget educational institution

Gymnasium No. 000

Design work on geometry.

Eight ways to construct a tangent to a circle.

9 biological and chemical class

supervisor: ,

Deputy Director for Academic Affairs,

mathematics teacher.

Moscow 2012

Introduction

Chapter 1. …………………………………………………………………4

Conclusion (conclusion)

Introduction

The highest manifestation of the spirit is the mind.

The highest manifestation of the mind is geometry.

The geometry cell is a triangle. He is the same

inexhaustible, like the universe. The circle is the soul of geometry.

Know the circumference and you will not only know the soul

geometry, but also exalt your soul.

Claudius Ptolemy
Task.

Construct a tangent to a circle with center O and radius R passing through a point A lying outside the circle

Chapter 1.

Constructions of a tangent to a circle that do not require justification based on the theory of parallel lines.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16 src=">ABO =90°. For a circle (O; r) OB - radius. OB AB, therefore, AB is a tangent on the basis of a tangent.

Similarly, AC is a tangent to a circle.

Construction No. 1 is based on the fact that the tangent of a circle is perpendicular to the radius drawn to the tangent point.

For a line there is only one point of contact with the circle.

Through a given point on a line, only one perpendicular line can be drawn.

Building number 2.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16"> ABO = 90°

5. OB - radius, ABO = 90°, therefore, AB - tangent on the basis.

6. Similarly, in an isosceles triangle AON, AC is a tangent (ACO \u003d 90 °, OS is a radius)

7. So, AB and AC are tangents

Building #3

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">OPM = OVA= 90° (as the corresponding angles in equal triangles), therefore, AB - tangent on the basis of a tangent.

4. Similarly, AC is a tangent

Building №4

https://pandia.ru/text/78/156/images/image008_9.jpg" align="left" width="330" height="743 src=">

Building number 6.

Building:

2. Draw through point A an arbitrary line that intersects the circle (O, r) at points M and N.

6. AB and BC are the desired tangents.

Proof:

1. Since triangles PQN and PQM are inscribed in a circle and the side PQ is the diameter of the circle, these triangles are right triangles.

2. In triangle PQL, segments PM and QN are heights intersecting at point K, so KL is the third height..gif" width="17" height="16 src=">.gif" width="17" height="16 src =">AQS =AMS = 180° - https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">PQN = β, then |AQ| = |AS|ctg β Therefore |PA| : |AQ| = ctg α: ctg β (2).

5. Comparing (1) and (2) we get |PD| : |PA| = |DQ| : |AQ|, or

(|OD| + R)(|OA| - R)=(R -|OD|)(|OA| + R).

After opening the brackets and simplifying, I find that |OD|·|OA|=R².

5. It follows from the relation |OD|·|OA|=R² that |OD|:R=R: |OA|, that is, triangles ODB and OBA are similar..gif" width="17" height="16"> OBA=90°. Therefore, the line AB is the required tangent, which was to be proved.

Building number 6.

Building:

1. Draw a circle (A; |OA|).

2. I will find a compass opening equal to 2R, for which I will choose a point S on the circle (O; R) and set aside three arcs containing 60º each: SP=PQ=QT=60°. Points S and T are diametrically opposed.

3. I build a circle (O; ST) intersecting w 1What is this circle? at points M and N.

4. Now I will build the middle MO. To do this, I build circles (O; OM) and (M; MO), and then for points M and O we find diametrically opposite points U and V on them.

6. Finally, I will construct a circle (K; KM) and (L; LM) intersecting at the desired point B - the middle of MO.

Proof:

Triangles KMV and UMK are isosceles and similar. Therefore, from the fact that KM \u003d 0.5MU, it follows that MB \u003d 0.5MK \u003d 0.5R. So, point B is the desired point of contact. Similarly, you can find the point of contact C.

Chapter 3

Construction of a tangent to a circle based on the properties of secants, bisectors.

Building #7

https://pandia.ru/text/78/156/images/image011_7.jpg" align="left" width="440" height="514 src="> Building #8

Building:

1. Construct a circle (A; AP) intersecting the line AP at point D.

2. Construct a circle w on the diameter QD

3. I will intersect it with a perpendicular to the line AR at point A and get points M and N.

Proof:

Obviously, AM²=AN²=AD·AQ=AP·AQ. Then the circle (A; AM) intersects (O; R) at the points of contact B and C. AB and AC are the desired tangents.