The concept of a tangent to a circle

The circle has three possible mutual arrangements relative to a straight line:

If the distance from the center of the circle to the line is less than the radius, then the line has two points of intersection with the circle.

If the distance from the center of the circle to the line is equal to the radius, then the line has two points of intersection with the circle.

If the distance from the center of the circle to the straight line is greater than the radius, then the straight line has two points of intersection with the circle.

We now introduce the concept of a tangent line to a circle.

Definition 1

A tangent to a circle is a straight line that has one point of intersection with it.

The common point of the circle and the tangent is called the tangent point (Fig. 1).

Figure 1. Tangent to a circle

Theorems related to the concept of a tangent to a circle

Theorem 1

Tangent property theorem: The tangent to the circle is perpendicular to the radius drawn to the tangent point.

Proof.

Consider a circle with center $O$. Let us draw the tangent $a$ at the point $A$. $OA=r$ (Fig. 2).

Let us prove that $a\bot r$

We will prove the theorem by the method of "by contradiction". Assume that the tangent $a$ is not perpendicular to the radius of the circle.

Figure 2. Illustration of Theorem 1

That is, $OA$ is oblique to a tangent. Since the perpendicular to the line $a$ is always less than the slope to the same line, the distance from the center of the circle to the line is less than the radius. As we know, in this case the line has two points of intersection with the circle. Which contradicts the definition of a tangent.

Therefore, the tangent is perpendicular to the radius of the circle.

The theorem has been proven.

Theorem 2

Converse to the tangent property theorem: If the line passing through the end of the radius of a circle is perpendicular to the radius, then this line is tangent to this circle.

Proof.

According to the condition of the problem, we have that the radius is a perpendicular drawn from the center of the circle to the given line. Therefore, the distance from the center of the circle to the straight line is equal to the length of the radius. As we know, in this case the circle has only one point of intersection with this line. By definition 1, we get that the given line is tangent to the circle.

The theorem has been proven.

Theorem 3

The segments of tangents to the circle, drawn from one point, are equal and make up equal angles with a straight line passing through this point and the center of the circle.

Proof.

Let there be given a circle centered at the point $O$. Two different tangents are drawn from the point $A$ (which lies on all circles). From the touch point $B$ and $C$ respectively (Fig. 3).

Let us prove that $\angle BAO=\angle CAO$ and that $AB=AC$.

Figure 3. Illustration of Theorem 3

By Theorem 1, we have:

Therefore, the triangles $ABO$ and $ACO$ are right triangles. Since $OB=OC=r$, and the hypotenuse $OA$ is common, these triangles are equal in hypotenuse and leg.

Hence we get that $\angle BAO=\angle CAO$ and $AB=AC$.

The theorem has been proven.

An example of a task on the concept of a tangent to a circle

Example 1

Given a circle with center $O$ and radius $r=3\ cm$. The tangent $AC$ has a tangent point $C$. $AO=4\cm$. Find $AC$.

Solution.

First, let's depict everything in the figure (Fig. 4).

Figure 4

Since $AC$ is a tangent and $OC$ is a radius, then by Theorem 1 we get $\angle ACO=(90)^(()^\circ )$. It turned out that the triangle $ACO$ is rectangular, which means, according to the Pythagorean theorem, we have:

\[(AC)^2=(AO)^2+r^2\] \[(AC)^2=16+9\] \[(AC)^2=25\] \

Let's draw CO and consider triangles OAC and OBC1) In ΔOAC and ΔOBC:OC - common, OA = OB, as radii, OA ⊥ CA, OB ⊥ CB (because AC and CB are tangents). Thus, ΔОAC = ΔOBC according to the 1st sign of equality of triangles. Whence AC = CO.2) Let three tangents to the circle be drawn through point C: CA, CB, CM. Then it follows that CA = CB = CM, whence the points A, B, M lie on the same circle with center C. It turned out that the two circles have three points in common. Contradiction. Circle theorem: Circles cannot intersect at more than two points. Thus, through given point it is impossible to draw more than two tangents to a given circle. Therefore, SA and CB are tangents to the circle and they are equal.

From point C we draw a segment CO. We get two triangles: ∆COA and ∆SOVV ∆SOA and ∆COV:CO - common, OA = OB, as radii, OA ⊥ CA, OV ⊥ CB (because CA and CB are tangents). Thus, ΔCOA = ΔCOV according to the 1st sign of the equality of triangles. Where SA = SW.

Related tasks:

1. In an arbitrary triangle, a middle line, which cuts off a smaller triangle from it. Find the ratio of the area of ​​the smaller triangle to the area of ​​the given triangle.

2. A circle is described around the trapezoid, the center of which is located on its larger base. Find the angles of a trapezoid if its smaller base is half the size of its larger base.

3. The angle between the bisector and the height drawn from the vertex of the larger angle of the triangle is 12*. Find the angles of this triangle if its largest angle is four times the smallest angle.

4. O1 and O2 are the centers of two tangents outwardly circles. Line O1O2 intersects the first circle (centered at point O1) at point A. Find the diameter of the second circle if the radius of the first is 5 cm, and the tangent drawn from point A to the second circle forms an angle of 30 * with line O1O2.

Direct ( MN) that has only one common point with the circle ( A), is called tangent to the circle.

The common point is called in this case touch point.

Possibility of existence tangent, and, moreover, drawn through any point circles, as a point of contact, is proved by the following theorem.

Let it be required to circles centered O tangent through a point A. For this, from the point A, as from the center, describe arc radius AO, and from the point O, as the center, we intersect this arc at points B and FROM compass solution equal to the diameter of the given circle.

After spending then chords OB and OS, connect the dot A with dots D and E where these chords intersect the given circle. Direct AD and AE - tangent to the circle O. Indeed, it is clear from the construction that triangles AOB and AOC isosceles(AO = AB = AC) with bases OB and OS, equal to the diameter of the circle O.

Because OD and OE are the radii, then D - middle OB, a E- middle OS, means AD and AE - medians drawn to the bases of isosceles triangles, and therefore perpendicular to these bases. If direct DA and EA perpendicular to the radii OD and OE, then they are tangents.

Consequence.

Two tangents drawn from the same point to the circle are equal and form equal angles with the line connecting this point with the center.

So AD=AE and ∠ OAD = ∠OAE because right triangles AOD and AOE having a common hypotenuse AO and equal legs OD and OE(as radii) are equal. Note that here the word “tangent” means the actual “ tangent segment” from the given point to the point of contact.

Most often, it is geometric problems that cause difficulties for applicants, graduates, and participants in mathematical Olympiads. If you look at the statistics of the USE in 2010, you can see that about 12% of the participants started the geometric task C4, and only 0.2% of the participants received a full score, and in general, the task turned out to be the most difficult of all proposed.

Obviously, the sooner we offer schoolchildren beautiful or unexpected in terms of the way they solve problems, the more likely they are to interest and captivate them seriously and for a long time. But how hard it is to find interesting and challenging tasks at the 7th grade level, when the systematic study of geometry is just beginning. What can be offered to a student interested in mathematics, who knows only the signs of the equality of triangles, the properties of adjacent and vertical angles? However, it is possible to introduce the concept of a tangent to a circle, as a straight line that has one common point with the circle; accept that the radius drawn to the point of contact is perpendicular to the tangent. Of course, it is worth considering all possible cases of the location of two circles and common tangents to them, which can be drawn from zero to four. By proving the theorems proposed below, it is possible to significantly expand the set of tasks for seventh graders. At the same time, along the way, to prove important or simply interesting and entertaining facts. Moreover, since many statements are not included in the school textbook, they can be discussed both in the classroom and with graduates when repeating planimetry. These facts turned out to be relevant in the last academic year. Since many diagnostic work itself USE work contained a problem for the solution of which it was necessary to use the property of a tangent segment proved below.

T 1 Segments of tangents to a circle drawn from
one point are equal (Fig. 1)

That's it with the theorem, you can first introduce the seventh graders.
In the process of proof, we used the sign of equality of right triangles, concluded that the center of the circle lies on the bisector of the angle BCA.
In passing, we remembered that the bisector of an angle is the locus of points of the inner region of the angle, equidistant from its sides. The solution of a far from trivial problem is based on these facts, accessible even to beginners in studying geometry.

1. Bisectors of angles BUT, AT and FROM convex quadrilateral ABCD intersect at one point. Rays AB and DC intersect at a point E, and the rays
sun and AD at the point F. Prove that a non-convex quadrilateral AECF the sum of the lengths of opposite sides are equal.

Solution (Fig. 2). Let O is the point of intersection of these bisectors. Then O equidistant from all sides of the quadrilateral ABCD, that is
is the center of a circle inscribed in a quadrilateral. By theorem 1 equalities are correct: AR = AK, ER = EP, FT = FK. We add the left and right parts term by term, we get the correct equality:

(AR + ER) + FT = (AK +FK) + EP; AE + (FC + CT) = AF + (EU + PC). Because ST = RS, then AE + FC = AF + EU, which was to be proved.

Let us consider a problem with an unusual formulation, for the solution of which it is sufficient to know the theorem 1 .

2. Is there n-gon whose sides are consecutively 1, 2, 3, ..., n in which the circle can be inscribed?

Solution. Let's say such n-gon exists. BUT 1 BUT 2 =1, …, BUT n-1 BUT n= n– 1,BUT n BUT 1 = n. B 1 , …, B n are the corresponding touch points. Then by Theorem 1 A 1 B 1 = A 1 B n< 1, n – 1 < A n B n< n. By the property of tangent segments A n B n= A n B n-1 . But, A n B n-1< A n-1 BUT n= n- 1. Contradiction. Therefore, no n-gon that satisfies the condition of the problem.

T 2 The sums of opposite sides of a quadrilateral circumscribed about
circles are equal (Fig. 3)

Schoolchildren, as a rule, easily prove this property of the described quadrilateral. After proving the theorem 1 , it is a training exercise. This fact can be generalized - the sums of the sides of the circumscribed even-gon, taken through one, are equal. For example, for a hexagon ABCDEF right: AB + CD + EF = BC + DE + FA.

3. Moscow State University. In a quadrilateral ABCD there are two circles: the first circle touches the sides AB, BC and AD, and the second - sides BC, CD and AD. On the sides BC and AD points are taken E and F accordingly, segment EF touches both circles, and the perimeter of the quadrilateral ABEF on the 2p greater than the perimeter of the quadrilateral ECDF. Find AB, if cd=a.

Solution (Fig. 1). Since the quadrilaterals ABEF and ECDF are inscribed, by Theorem 2 Р ABEF = 2(AB + EF) and Р ECDF = 2(CD + EF), by condition

P ABEF - P ECDF = 2(AB + EF) - 2(CD + EF) = 2p. AB-CD=p. AB = a + p.

Core task 1. Direct AB and AC are tangents at points AT and FROM to a circle centered at point O. Through an arbitrary point X arcs sun
a tangent to a circle is drawn that intersects the segments AB and AC at points M and R respectively. Prove that the perimeter of the triangle USAID and the angle MPA do not depend on the choice of point X.

Solution (Fig. 5). By Theorem 1 MB = MX and PC = RX. So the perimeter of the triangle USAID equal to the sum of the segments AB and AS. Or double tangent drawn to the excircle for a triangle USAID . The value of the MOP angle is measured by half the value of the angle WOS, which does not depend on the choice of point X.

Reference task 2a. In a triangle with sides a, b and c inscribed circle tangent to side AB and point TO. Find the length of a segment AK.

Solution (Fig. 6). Method one (algebraic). Let AK \u003d AN \u003d x, then BK = BM = c - x, CM = CN = a - c + x. AC = AN + NC, then we can write an equation for x: b \u003d x + (a - c + x). Where .

Method two (geometric). Let's turn to the diagram. Segments of equal tangents, taken one at a time, add up to a semi-perimeter
triangle. Red and green make up a side a. Then the segment of interest to us x = p - a. Of course, the results obtained are consistent.

Supporting task 2b. Find the length of the tangent segment ak, if To is the point of tangency of the excircle with the side AB. Solution (Fig. 7). AK = AM = x, then BK = BN = c - x, CM = CN. We have the equation b + x = a + (c - x). Where . Z Note that from the basic problem 1 follows that CM = p ∆ ABC. b+x=p; x \u003d p - b. The obtained formulas are used in the following tasks.

4. Find the radius of a circle inscribed in a right triangle with legs a, b and hypotenuse With. Solution (Fig. 8). T how OMCN- square, then the radius of the inscribed circle is equal to the segment of the tangent CN. .

5. Prove that the points of tangency of the inscribed and excircle circles with the side of the triangle are symmetrical with respect to the midpoint of this side.

Solution (Fig. 9). Note that AK is the segment of the tangent of the excircle for the triangle ABC. By formula (2) . VM- line segment tangent incircle for triangle ABC. According to formula (1) . AK = VM, and this means that the points K and M equidistant from the middle of the side AB, Q.E.D.

6. Two common outer tangents and one inner tangent are drawn to two circles. The inner tangent intersects the outer ones at points A, B and touches circles at points A 1 and IN 1 . Prove that AA 1 \u003d BB 1.

Solution (Fig. 10). Stop ... But what is there to decide? It's just another formulation of the previous problem. It is obvious that one of the circles is inscribed and the other is excircle for some triangle ABC. And the segments AA 1 and BB 1 correspond to segments AK and VM tasks 5. It is noteworthy that the problem proposed at the All-Russian Olympiad for schoolchildren in mathematics is solved in such an obvious way.

7. The sides of the pentagon are 5, 6, 10, 7, 8 in the order of going around. Prove that a circle cannot be inscribed in this pentagon.

Solution (Fig. 11). Let's assume that the pentagon ABCDE you can inscribe a circle. Moreover, the parties AB, BC, CD, DE and EA are equal to 5, 6, 10, 7 and 8, respectively. F, G, H, M and N. Let the length of the segment AF is equal to X.

Then bf = FD – AF = 5 – x = BG. GC = BC – BG = = 6 – (5 – x) = 1 + x = CH. And so on: HD = DM = 9 – x; ME = EN = x – 2, AN = 10 – X.

But, AF = AN. That is 10 - X = X; X= 5. However, the segment of the tangent AF cannot equal side AB. The resulting contradiction proves that a circle cannot be inscribed in a given pentagon.

8. A circle is inscribed in a hexagon, its sides in the bypass order are 1, 2, 3, 4, 5. Find the length of the sixth side.

Solution. Of course, the tangent segment can be denoted as X, as in the previous problem, write an equation and get an answer. But, it is much more efficient and effective to use the note to the theorem 2 : the sums of the sides of the circumscribed hexagon, taken through one, are equal.

Then 1 + 3 + 5 = 2 + 4 + X, where X- unknown sixth side, X = 3.

9. Moscow State University, 2003. Faculty of Chemistry, No. 6(6). into a pentagon ABCDE inscribed circle, R is the point of contact of this circle with the side sun. Find the length of the segment VR, if it is known that the lengths of all sides of the pentagon are whole numbers, AB = 1, CD = 3.

Solution (fig.12). Since the lengths of all sides are integers, the fractional parts of the lengths of the segments are equal BT, BP, DM, DN, AK and AT. We have AT + TV= 1, and the fractional parts of the lengths of the segments AT and TV are equal. This is possible only when AT + TV= 0.5. By theorem 1 WT + VR.
Means, VR= 0.5. Note that the condition CD= 3 turned out to be unclaimed. Obviously, the authors of the problem assumed some other solution. Answer: 0.5.

10. In a quadrilateral ABCD AD=DC, AB=3, BC=5. Circles inscribed in triangles ABD and CBD touch the segment BD at points M and N respectively. Find the length of a segment MN.

Solution (Fig. 13). MN = DN - DM. According to formula (1) for triangles DBA and DBC respectively, we have:

11. In a quadrilateral ABCD you can inscribe a circle. Circles inscribed in triangles ABD and CBD have radii R and r respectively. Find the distance between the centers of these circles.

Solution (Fig. 13). Since, by the condition, the quadrilateral ABCD inscribed, by theorem 2 we have: AB + DC = AD + BC. Let's use the idea of ​​solving the previous problem. . This means that the points of contact of the circles with the segment DM match. The distance between the centers of the circles is equal to the sum of the radii. Answer: R + r.

In fact, it is proved that the condition is in a quadrilateral ABCD you can inscribe a circle, which is equivalent to the condition - in a convex quadrilateral ABCD circles inscribed in triangles ABC and ADC touch each other. The opposite is true.

It is proposed to prove these two mutually inverse statements in the following problem, which can be considered a generalization of this one.

12. In a convex quadrilateral ABCD (rice. fourteen) circles inscribed in triangles ABC and ADC touch each other. Prove that circles inscribed in triangles ABD and bdc also touch each other.

13. In a triangle ABC with the parties a, b and c on the side sun marked point D so that the circles inscribed in the triangles ABD and ACD touch the segment AD at one point. Find the length of a segment BD.

Solution (Fig. 15). We apply formula (1) for triangles ADC and adb, calculating DM two

Turns out, D- point of contact with the side sun circle inscribed in a triangle ABC. The opposite is true: if the vertex of the triangle is connected to the tangent point of the inscribed circle on opposite side, then the circles inscribed in the resulting triangles touch each other.

14. Centers O 1 , O 2 and O 3 three non-intersecting circles of the same radius are located at the vertices of the triangle. From points O 1 , O 2 , O 3, tangents to these circles are drawn as shown in the figure.

It is known that these tangents, intersecting, formed a convex hexagon, the sides of which through one are colored red and blue. Prove that the sum of the lengths of the red segments is equal to the sum of the lengths of the blue ones.

Solution (Fig. 16). It is important to understand how to use the fact that given circles have the same radii. Note that the segments BR and DM are equal, which follows from the equality of right triangles O 1 BR and O 2 BM. Similarly DL = D.P., FN = FK. We add the equalities term by term, then subtract from the resulting sums the same segments of tangents drawn from the vertices BUT, FROM, and E hexagon ABCDEF: AR and AK, CL and CM, EN and EP. We get what we need.

Here is an example of a stereometry problem proposed at the XII International Mathematical Tournament for High School Students “A. N. Kolmogorov Memory Cup”.

16. Given a pentagonal pyramid SA 1 A 2 A 3 A 4 A 5 . There is a scope w, which touches all the edges of the pyramid and another sphere w 1 , that touches all sides of the base A 1 A 2 A 3 A 4 A 5 and extensions of the lateral ribs SA 1 , SA 2 , SA 3 , SA 4 , SA 5 for the tops of the base. Prove that the vertex of the pyramid is equidistant from the vertices of the base. (Berlov S. L., Karpov D. V.)

Solution. The intersection of the sphere w with the plane of any of the sphere's faces is the inscribed circle of the face. The intersection of the sphere w 1 with each of the faces SA i A i+1 - excircle tangent to side A i A i+1 triangle SA i A i+1 and continuations of the other two sides. Denote the point of contact w 1 with the extension of the side SA i through B i. By reference problem 1, we have that SB i = SB i +1 = p SAiAi+1 , therefore, the perimeters of all side faces of the pyramid are equal. Denote the tangent point w with the side SA i through C i. Then SC 1 = SC 2 = SC 3 = SC 4 = SC 5 = s,
since the segments of the tangents are equal. Let C i A i = a i. Then p SAiAi +1 = s+a i +a i+1 , and it follows from the equality of perimeters that a 1 = a 3 = a 5 = a 2 = a 4 , whence SA 1 = SA 2 = SA 3 = SA 4 = SA 5 .

17. USE. Diagnostic work December 8, 2009, С–4. Dana trapezoid ABCD, the bases of which BC= 44,AD = 100, AB=CD= 35. Circle tangent to lines AD and AC touches the side CD at the point K. Find the length of the segment CK.VDC and BDA, touch the side BD at points E and F. Find the length of the segment EF.

Solution. Two cases are possible (Fig. 20 and Fig. 21). Using formula (1), we find the lengths of the segments DE and D.F..

In the first case AD = 0,1AC, CD = 0,9AC. In the second - AD = 0,125AC, CD = 1,125AC. We substitute the data and get the answer: 4.6 or 5.5.

Tasks for independent solution /

1. The perimeter of an isosceles trapezoid inscribed about a circle is 2r. Find the projection of the diagonal of the trapezoid onto the larger base. (1/2p)

2. Open bank USE tasks mathematics. AT 4. To a circle inscribed in a triangle ABC (fig. 22), three tangents are drawn. The perimeters of the truncated triangles are 6, 8, 10. Find the perimeter of this triangle. (24)

3. Into a triangle ABC inscribed circle. MN- tangent to the circle MО AC, NО BC, BC = 13, AC = 14, AB = 15. Find the perimeter of a triangle MNC. (12)

4. To a circle inscribed in a square with side a, a tangent is drawn that intersects two of its sides. Find the perimeter of the cut off triangle. (a)

5. A circle is inscribed in a pentagon with sides a, d, c, d and e. Find the segments into which the point of contact divides the side equal to a.

6. A circle is inscribed in a triangle with sides 6, 10 and 12. A tangent is drawn to the circle so that it intersects two large sides. Find the perimeter of the cut off triangle. (16)

7. CD is the median of the triangle ABC. Circles inscribed in triangles ACD and BCD, touch the segment CD at points M and N. Find MN, if AC – sun = 2. (1)

8. In a triangle ABC with the parties a, b and c on the side sun marked point D. To circles inscribed in triangles ABD and ACD, a common tangent is drawn that intersects AD at the point M. Find the length of a segment AM. (Length AM does not depend on the position of the point D and
equals ½ ( c + b - a))

9. A circle of radius is inscribed in a right triangle a. The radius of the circle tangent to the hypotenuse and the extensions of the legs is R. Find the length of the hypotenuse. ( R-a)

10. In a triangle ABC the lengths of the sides are known: AB = With, AC = b, sun = a. A circle inscribed in a triangle is tangent to a side AB at the point From 1. The excircle is tangent to the extension of the side AB per point BUT at the point From 2. Determine the length of the segment S 1 S 2. (b)

11. Find the lengths of the sides of the triangle, divided by the point of contact of the inscribed circle of radius 3 cm into segments 4 cm and 3 cm. (7, 24 and 25 cm in a right triangle)

12. Soros Olympiad 1996, 2nd round, 11th grade. Triangle given ABC, on the sides of which points are marked A 1, B 1, C 1. Radii of circles inscribed in triangles AC 1 B 1 , BC 1 A 1 , CA 1 B 1 equal in r. Radius of a circle inscribed in a triangle A 1 B 1 C 1 equals R. Find the radius of a circle inscribed in a triangle ABC. (R +r).

Problems 4–8 are taken from R. K. Gordin’s problem book “Geometry. Planimetry." Moscow. Publishing house MTSNMO. 2004.

1. Two tangents from one point.

Let two tangents $$AM$$ and $$AN$$ be drawn to the circle centered at the point $$O$$, the points $$M$$ and $$N$$ lie on the circle (Fig. 1).

By the definition of tangent $$OM \perp AM$$ and $$ON \perp AN$$. In right-angled triangles $$AOM$$ and $$AON$$ the hypotenuse $$AO$$ is common, the legs of $$OM$$ and $$ON$$ are equal, so $$\Delta AOM = \Delta AON$$. The equality of these triangles implies $$AM=AN$$ and $$\angle MAO = \angle NAO$$. Thus, if two tangents are drawn from a point to a circle, then:

1.1$$(\^{\circ}$$. !} the segments of the tangents from this point to the points of contact are equal;

1.2$$(\^{\circ}$$. !} a straight line passing through the center of the circle and a given point bisects the angle between the tangents.

Using property 1.1$$(\^{\circ}$$, легко решим следующие две задачи. (В решении используется тот факт, что в каждый треугольник можно вписать окружность).!}

Point $$D$$ is located on the basis $$AC$$ of the isosceles triangle $$ABC$$, while $$DA = a$$, $$DC = b$$ (Fig. 2). Circles inscribed in triangles $$ABD$$ and $$DBC$$ touch line $$BD$$ at points $$M$$ and $$N$$ respectively. Find the segment $$MN$$.

.

$$\triangle$$ Let $$a > b $$. Denote $$x = MN$$, $$y = ND$$, $$z = BM$$.

By the property of tangents $$DE = y$$, $$KD = x + y $$, $$AK = AP = a - (x + y)$$, $$CE = CF = b - y$$, $ $BP = z$$, and $$BF = z + x$$. Let's express the sides (Fig. 2a): $$AB = z+a-x-y$$, $$BC=z+x-b-y$$. By condition $$AB=BC$$, so $$z+a-x -y = z+x+b-y$$. From here we find $$x=\frac((a-b))(2)$$, i.e. $$MN=\frac((a-b))(2)$$. If $$a \lt b$$, then $$MN=\frac((b-a))(2)$$. So $$MN=\frac(1)(2)|a-b|$$. $$\blacktriangle$$

ANSWER

$$\frac(|a-b|) (2)$$

Prove that in a right triangle the sum of the legs is equal to twice the sum of the radii of the inscribed and circumscribed circles, i.e. $$a+b=2R+2r$$.

$$\triangle$$ Let $$M$$, $$N$$ and $$K$$ be the points where the circle touches the sides right triangle$$ABC$$ (Fig. 3), $$AC=b$$, $$BC=a$$, $$r$$ is the radius of the inscribed circle, $$R$$ is the radius of the circumscribed circle. Recall that the hypotenuse is the diameter of the circumscribed circle: $$AB=2R$$. Further, $$OM \perp AC$$, $$BC \perp AC$$, so $$OM \parallel BC$$, similar to $$ON \perp BC$$, $$AC \perp BC$$, so $$ON \parallel AC$$. The quadrilateral $$MONC$$ is by definition a square, all its sides are $$r$$, so $$AM = b - r$$ and $$BN = a - r $$.

By the property of the tangents $$AK=AM$$ and $$BK=BN$$, therefore $$AB = AK + KB = a+b-2r$$, and since $$AB=2R$$ , then we get $$a+b=2R+2r$$. $$\blacktriangle$$

Property 1.2$$(\^{\circ}$$ сформулируем по другому: !} The center of a circle inscribed in an angle lies on the bisector of that angle.

A trapezoid $$ABCD$$ with bases $$AD$$ and $$BC$$ is circumscribed near a circle centered at $$O$$ (Fig. 4a).

a) Prove that $$\angle AOB = \angle COD = $$90$$(\^{\circ}$$ .!}

b) Find the radius of the circle if $$BO = \sqrt(5)$$ and $$AO = 2 \sqrt(5)$$. (Fig. 4b)

$$\triangle$$ a) The circle is inscribed in the angle $$BAD$$, by property 1.2$$(\^{\circ}$$ $$AO$$ - биссектриса угла $$A$$, $$\angle 1 = \angle 2 = \frac{1}{2} \angle A$$; $$BO$$ - биссектриса угла $$B$$, $$\angle 3 = \angle 4 = \frac{1}{2} \angle B$$. Из параллельности прямых $$AD$$ и $$BC$$ следует, что $$\angle A + \angle B = 180^{\circ}$$,поэтому в треугольнике $$AOB$$ из $$\angle 1 + \angle 3 = \frac{1}{2} (\angle A + \angle B) = 90^{\circ}$$ следует $$\angle AOB = 90^{\circ}$$.!}

Similarly $$CO$$ and $$DO$$ are the bisectors of angles $$C$$ and $$D$$ of a trapezoid, $$\angle COD = 180^(\circ) - \frac(1)(2)(\ angle C + \angle D) = 90^(\circ)$$.

b) The triangle $$AOB$$ is a right triangle with legs $$AO = 2 \sqrt(5)$$ and $$BO = \sqrt(5)$$. Find the hypotenuse $$AB=\sqrt(20+5) = 5$$. If the circle is tangent to side $$AB$$ at point $$K$$, then $$OK \perp AB$$ and $$OK$$ are the radius of the circle. By the right triangle property $$AB \cdot OK = AO \cdot BO$$, whence $$OK = \frac(2\sqrt(5)\cdot \sqrt(5))(5) = 2$$. $$\blacktriangle$$

ANSWER

2. Angle between a tangent and a chord with a common point on a circle.

Recall that the degree measure of an inscribed angle is half degree measure arc on which it rests.

Theorem 1. The measure of the angle between the tangent and the chord, having a common point on the circle, is equal to half the degree measure of the arc enclosed between its sides.

$$\square$$ Let $$O$$ be the center of the circle, $$AN$$ be the tangent (Fig. 5). The angle between the tangent $$AN$$ and the chord $$AB$$ is denoted by $$\alpha$$. Connect the points $$A$$ and $$B$$ to the center of the circle.

Thus, the degree measure of the angle between the tangent and the chord is equal to half the degree measure of the arc $$AnB$$, which is enclosed between its sides, and, therefore, the angle $$BAN$$ is equal to any inscribed angle based on the arc $$AnB$$ . (Similar reasoning can be carried out for the angle $$MAB$$). $$\blacksquare$$

The point $$C$$ lies on the circle and is separated from the tangents drawn from the point $$M$$ to the circle at a distance $$CS = a$$ and $$CP = b$$ (Fig. 6). Prove that $$CK = \sqrt(ab)$$.

$$\triangle$$ Let's draw the chords $$CA$$ and $$CB$$. The angle $$SAC$$ between the tangent $$SA$$ and the chord $$AC$$ is equal to the inscribed angle $$ABC$$. And the angle $$PBC$$ between the tangent $$PB$$ and the chord $$BC$$ is equal to the inscribed angle $$BAC$$. We got two pairs of similar right triangles $$\Delta ASC \sim\Delta BKC$$ and $$\Delta BPC \sim \Delta AKC$$. From the similarity, we have $$\dfrac(a)(AC)=\dfrac(x)(BC)$$ and $$\dfrac(b)(BC)=\dfrac(x)(AC)$$, which implies $ $ab=x^2$$, $$x=\sqrt(ab)$$. (If the projection of the point $$C$$ onto the line $$AB$$ lies outside the segment $$AB$$, the proof does not change much). (H. etc.) $$\blacktriangle$$

Reception, applied in the solution - drawing "missing" chords - often helps in problems and theorems with a circle and a tangent, as, for example, in the proof of the following theorem "about tangent and secant".

Theorem 2. If a tangent $$MA$$ and a secant $$MB$$ are drawn to the circle from the same point $$M$$ and intersect the circle at the point $$C$$ (Fig. 7), then $$MA ^2 = MB \cdot MC$$, i.e. if a tangent and a secant are drawn from the point $$M$$ to the circle, then the square of the segment of the tangent from the point $$M$$ to the point of tangency is equal to the product of the lengths of the segments of the secant from the point $$M$$ to the points of its intersection with the circle.

$$\square$$ Let's draw the chords $$AC$$ and $$AB$$. The angle $$MAC$$ between the tangent and the chord is equal to the inscribed angle $$ABC$$, both measured by half the degree measure of the arc $$AnC$$. In triangles $$MAC$$ and $$MBA$$ the angles $$MAC$$ and $$MBA$$ are equal, and the vertex angle $$M$$ is common. These triangles are
are good, from the similarity we have $$MA/MB = MC/MA$$, which implies $$MA^2 = MB \cdot MC$$. $$\blacksquare$$

The radius of the circle is $$R$$. A tangent $$MA$$ and a secant $$MB$$ passing through the center $$O$$ of the circle are drawn from the point $$M$$ (Fig. 8). Find the distance between the point $$M$$ and the center of the circle if $$MB = 2MA$$.

$$\triangle$$ (x+R)/2$$. By the tangent and secant theorem $$(x+R)^2/4=(x+R)(x-R)$$, whence, canceling by $$(x+R)$$, we get $$(x+R )/4=x-R$$. We easily find $$x = \dfrac(5)(3)R$$. $$\blacktriangle$$

ANSWER

$$\dfrac(5)(3)R$$

3. Property of chords of a circle.

It is useful to prove these properties on your own (it is better fixed), you can analyze the proofs from the textbook.

1.3$$(\^{\circ}$$. Диаметр, перпендикулярный хорде, делит её пополам. Обратно: диаметр, проходящей через середину хорды (не являющуюся диаметром) перпендикулярен ей. !}

1.4$$(\^{\circ}$$. Равные хорды окружности находятся на !} equal distance from the center of the circle. Conversely, there are equal chords at an equal distance from the center of the circle.

1.5$$(\^{\circ}$$. !} The arcs of a circle enclosed between parallel chords are equal (Fig. 9 will tell you the way of the proof).

1.6$$(\^{\circ}$$. Если две хорды $$AB$$ и $$CD$$ пересекаются в точке $$M$$, то $$AM \cdot MB = CM \cdot MD$$, т. е. произведение длин отрезков одной хорды равно произведению длин отрезков другой хорды (на рис. 10 $$\Delta AMC \sim \Delta DMB$$). !}

We will prove the following assertion.

1.7$$(\^{\circ}$$. !} If in a circle of radius $$R$$ the inscribed angle based on a chord of length $$a$$ is equal to $$\alpha$$, then $$a = 2R\textrm(sin)\alpha$$.

$$\blacksquare$$ Let the chord $$BC = a$$ in a circle of radius $$R$$, the inscribed angle $$BAC$$ rest on the chord $$a$$, $$\angle BAC = \alpha$$ (Fig. 11 a, b).

Draw the diameter $$BA^(")$$ and consider the right triangle $$BA^(")C$$ ($$\angle BCA^(")= 90^(\circ)$$, based on the diameter).

If the angle $$A$$ is acute (Fig. 11a), then the center $$O$$ and the vertex $$A$$ lie on the same side of the line $$BC$$, $$\angle A^(") = \angle A$$ and $$BC = BA^(") \cdot \textrm(sin)A^(")$$, i.e. $$a=2R\textrm(sin)A^(")$ $ .

If angle $$A$$ is obtuse, center $$O$$ and vertex $$A$$ lie on opposite sides of line $$BC$$ (Fig. 11b), then $$\angle A^(") = 180^(\circ) - \angle A$$ and $$BC = BA^(") \cdot \textrm(sin)A^(")$$, i.e. $$a=2R\textrm(sin )(180-A^("))=2R\textrm(sin)A^(")$$.

If $$\alpha = 90^(\circ)$$, then $$BC$$ is the diameter, $$BC = 2R = 2R\textrm(sin)90^(\circ)$$.

In all cases, $$a=2R\textrm(sin)A^(")$$ . $$\blacktriangle$$

So $$\boxed(a = 2R\textrm(sin)\alpha)$$ or $$\boxed(R = \dfrac(a)(2\textrm(sin)\alpha))$$. (*)

Find the radius of a circle circumscribed about triangle $$ABC$$ where $$AB = 3\sqrt(3)$$, $$BC = 2$$ and angle $$ABC = 150^(\circ)$$.

$$\triangle$$ In the circle circumscribed about the triangle $$ABC$$, the angle $$B$$ is known, based on the chord $$AC$$. The above formula implies $$R = \dfrac(AC)(2\textrm(sin)B)$$.

Let us apply the cosine theorem to the triangle $$ABC$$ (Fig. 12) while taking into account that

$$\textrm(cos)150^(\circ) = \textrm(cos)(180^(\circ)-30^(\circ)) = -\textrm(cos)30^(\circ) = -\ dfrac(\sqrt(3))(2)$$, we get

$$AC^2 = 27+4+2\cdot 3\sqrt(3) \cdot 2 \cdot \dfrac(\sqrt(3))(2) = 49,\: AC=7$$.

Find $$R = \dfrac(AC)(2\textrm(sin)150^(\circ)) = \dfrac(7)(2\textrm(sin)30^(\circ)) = 7$$. $$\blacktriangle$$

ANSWER

We use the property of intersecting chords to prove the following theorem.

Theorem 3. Let $$AD$$ be the bisector of triangle $$ABC$$, then

$$AD^2 = AB\cdot AC - BD\cdot CD$$ , i.e. if$$AB=c,\: AC=b,\: BD=x,\:DC=y$$ , then$$AD^2 = bc-xy$$ (Fig. 13a).

$$\square$$ Let us describe a circle around the triangle $$ABC$$ (Fig. 13b) and denote the intersection point of the continuation of the bisector $$AD$$ with the circle as $$B_1$$. Denote $$AD = l $$ and $$DB_1 = z $$. Inscribed angles $$ABC$$ and $$AB_1C$$ are equal, $$AD$$ is the bisector of angle $$A$$, so $$\Delta ABD \sim \Delta AB_1C$$ (on two angles). From the similarity we have $$\dfrac(AD)(AC) = \dfrac(AB)(AB_1)$$, i.e. $$\dfrac(l)(b) = \dfrac(c)(l+z) $$, whence $$l^2=bc-lz$$. By the property of intersecting chords $$BD\cdot DC = AD \cdot DB_1$$, i.e. $$xy=lz$$, so we get $$l^2=bc-xy$$ . $$\blacksquare$$

4. Two touching circles

To conclude this section, consider problems with two tangent circles. Two circles that have a common point and a common tangent at that point are called tangent. If the circles are located on the same side of a common tangent, they are called related internally(Fig. 14a), and if located on opposite sides of the tangent, then they are called externally related(Fig. 14b).

If $$O_1$$ and $$O_2$$ are the centers of the circles, then by the definition of the tangent $$AO_1 \perp l$$, $$AO_2 \perp l$$, therefore, in both cases common pointtouch lies on the line of centers.

Two circles of radii $$R_1$$ and $$R_2$$ ($$R_1 > R_2$$) touch internally at point $$A$$. A line is drawn through the point $$B$$ lying on the larger circle and tangent to the smaller circle at the point $$C$$ (Fig. 15). Find $$AB$$ if $$BC = a$$.

$$\triangle$$ Let $$O_1$$ and $$O_2$$ be the centers of the larger and smaller circles, $$D$$ be the intersection point of the chord $$AB$$ with the smaller circle. If $$O_1N \perp AB$$ and $$O_2M \perp AB$$, then $$AN=AB/2$$ and $$AM=AD/2$$ (because the radius perpendicular to the chord divides cut it in half). The similarity of triangles $$AO_2M$$ and $$AO_1N$$ implies $$AN:AM = AO_1:AO_2$$ and hence $$AB:AD = R_1:R_2$$.

According to the tangent and secant theorem, we have:

$$BC^2 = AB\cdot BD = AB (AB-AD) = AB^2(1 - \dfrac(AD)(AB))$$,

i.e. $$a^2 = AB^2(1-\dfrac(R_2)(R_1))$$.

So $$AB = a \sqrt(\dfrac(R_1)(R_1-R_2))$$. $$\blacktriangle$$

Two circles of radii $$R_1$$ and $$R_2$$ touch externally at the point $$A$$ (Fig. 16). Their common outer tangent touches the larger circle at $$B$$ and the smaller one at $$C$$. Find the radius of the circle circumscribed about the triangle $$ABC$$.

$$\triangle$$ Connect the centers $$O_1$$ and $$O_2$$ with the points $$B$$ and $$C$$. By definition of tangent, $$O_1B \perp BC$$ and $$O_2C \perp BC$$. Hence $$O_1B \parallel O_2C$$ and $$\angle BO_1O_2 + \angle CO_2O_1 = 180^(\circ)$$. Since $$\angle ABC = \dfrac(1)(2) \angle BO_1A$$ and $$\angle ACB = \dfrac(1)(2) \angle CO_2A$$, then $$\angle ABC + \ angle ACB = 90^(\circ)$$. It follows that $$\angle BAC = 90^(\circ)$$ , and therefore the radius of the circle circumscribed about the right triangle $$ABC$$ is equal to half of the hypotenuse $$BC$$.

Let's find $$BC$$. Let $$O_2K \perp O_1B$$, then $$KO_2 = BC,\: O_1K = R_1-R_2,\: O_1O_2 = R_1+R_2$$. By the Pythagorean theorem we find:

$$KO_2 = \sqrt(O_1O_2^2 - O_1K^2)= 2\sqrt(R_1R_2), \: \underline(BC = 2\sqrt(R_1R_2) )$$.

So, the radius of the circumscribed triangle $$ABC$$ is equal to $$\sqrt(R_1R_2)$$. In the solution $$R_1 > R_2$$, for $$R_1

ANSWER

$$\sqrt(R_1R_2)$$