Tangent to a circle. Complete lessons - Knowledge Hypermarket. Tangent line What is a tangent angle

1. The distance from the center of the circle to the straight line is greater than the radius. In this case, all points of the line lie outside the circle.


2. The distance from the center of the circle to the straight line is less than the radius. In this case, the straight line has points lying inside the circle and since the straight line is infinite in both directions, it is intersected by the circle at 2 points.


3. The distance from the center of the circle to the straight line is equal to the radius. Straight line is tangent.

A straight line that has only one point in common with a circle is called tangent to the circle.

The common point is called in this case point of contact.

The possibility of the existence of a tangent, and, moreover, drawn through any point of the circle as a point of tangency, is proved by the following theorem.

Theorem. If a line is perpendicular to the radius at its end lying on the circle, then this line is a tangent.

Let O (fig) be the center of some circle and OA some of its radius. Through its end A we draw MN ^ OA.

It is required to prove that the line MN is tangent, i.e. that this line has only one common point A with the circle.

Let us assume the opposite: let MN have another common point with the circle, for example B.

Then straight line OB would be a radius and therefore equal to OA.

But this cannot be, since if OA is perpendicular, then OB must be inclined to MN, and the inclined one is greater than the perpendicular.

Converse theorem. If a line is tangent to a circle, then the radius drawn to the point of tangency is perpendicular to it.

Let MN be the tangent to the circle, A the point of tangency, and O the center of the circle.

It is required to prove that OA^MN.

Let's assume the opposite, i.e. Let us assume that the perpendicular dropped from O to MN will not be OA, but some other line, for example, OB.

Let's take BC = AB and carry out OS.

Then OA and OS will be inclined, equally distant from the perpendicular OB, and therefore OS = OA.

It follows from this that the circle, taking into account our assumption, will have two common points with the line MN: A and C, i.e. MN will not be a tangent, but a secant, which contradicts the condition.

Consequence. Through any given point on a circle one can draw a tangent to this circle, and only one, since through this point one can draw a perpendicular, and only one, to the radius drawn into it.

Theorem. A tangent parallel to a chord divides the arc subtended by the chord in half at the point of contact.

Let straight line AB (fig.) touch the circle at point M and be parallel to chord CD.

We need to prove that ÈCM = ÈMD.

Drawing the diameter ME through the point of tangency, we obtain: EM ^ AB, and therefore EM ^ CB.

Therefore CM=MD.

Task. Through a given point draw a tangent to a given circle.

If a given point is on a circle, then draw a radius through it and a perpendicular straight line through the end of the radius. This line will be the desired tangent.

Let us consider the case when the point is given outside the circle.

Let it be required (Fig.) to draw a tangent to a circle with center O through point A.

To do this, from point A, as the center, we describe an arc with radius AO, and from point O, as the center, we intersect this arc at points B and C with a compass opening equal to the diameter of the given circle.

Having then drawn the chords OB and OS, we connect point A with points D and E, at which these chords intersect with the given circle.

Lines AD and AE are tangent to circle O.

Indeed, from the construction it is clear that the pipes AOB and AOC are isosceles (AO = AB = AC) with the bases OB and OS equal to the diameter of the circle O.

Since OD and OE are radii, then D is the middle of OB, and E is the middle of OS, which means AD and AE are medians drawn to the bases of isosceles pipes, and therefore perpendicular to these bases. If the lines DA and EA are perpendicular to the radii OD and OE, then they are tangent.

Consequence. Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.

So AD=AE and ÐOAD = ÐOAE (Fig.), because rectangular tr-ki 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 a given point to the point of contact.

Task. Draw a tangent to a given circle O parallel to a given straight line AB (Fig.).

We lower a perpendicular OS to AB from the center O and through the point D, at which this perpendicular intersects the circle, draw EF || AB.

The tangent we are looking for will be EF.

Indeed, since OS ^ AB and EF || AB, then EF ^ OD, and the line perpendicular to the radius at its end lying on the circle is a tangent.

Task. Draw a common tangent to two circles O and O 1 (Fig.).

Analysis. Let's assume that the problem is solved.

Let AB be the common tangent, A and B the points of tangency.

Obviously, if we find one of these points, for example, A, then we can easily find the other one.

Let us draw the radii OA and O 1 B. These radii, being perpendicular to the common tangent, are parallel to each other.

Therefore, if from O 1 we draw O 1 C || BA, then the pipeline OCO 1 will be rectangular at vertex C.

As a result, if we describe a circle from O as the center with radius OS, then it will touch the straight line O 1 C at point C.

The radius of this auxiliary circle is known: it is equal to OA – CA = OA - O 1 B, i.e. it is equal to the difference between the radii of these circles.

Construction. From the center O we describe a circle with a radius equal to the difference of these radii.

From O 1 we draw a tangent O 1 C to this circle (in the manner indicated in the previous problem).

Through the tangent point C we draw the radius OS and continue it until it meets the given circle at point A. Finally, from A we draw AB parallel to CO 1.

In exactly the same way we can construct another common tangent A 1 B 1 (Fig.). Direct lines AB and A 1 B 1 are called external common tangents.

You can spend two more internal tangents as follows:

Analysis. Let's assume that the problem is solved (Fig.). Let AB be the desired tangent.

Let us draw the radii OA and O 1 B to the tangent points A and B. Since these radii are both perpendicular to the common tangent, they are parallel to each other.

Therefore, if from O 1 we draw O 1 C || BA and continue OA to point C, then OS will be perpendicular to O 1 C.

As a result, the circle described by the radius OS from point O as the center will touch the straight line O 1 C at point C.

The radius of this auxiliary circle is known: it is equal to OA+AC = OA+O 1 B, i.e. it is equal to the sum of the radii of the given circles.

Construction. From O as the center, we describe a circle with a radius equal to the sum of these radii.

From O 1 we draw a tangent O 1 C to this circle.

We connect the point of contact C with O.

Finally, through point A, at which OS intersects the given circle, we draw AB = O 1 C.

In a similar way we can construct another internal tangent A 1 B 1.

General definition of tangent

Let a tangent AT and some secant AM be drawn through point A to a circle with a center (Fig.).

Let's rotate this secant around point A so that the other intersection point B moves closer and closer to A.

Then the perpendicular OD, lowered from the center to the secant, will approach the radius OA more and more, and the angle AOD may become less than any small angle.

The angle MAT formed by the secant and tangent is equal to the angle AOD (due to the perpendicularity of their sides).

Therefore, as point B approaches A indefinitely, angle MAT can also become arbitrarily small.

This is expressed in other words like this:

a tangent is the limiting position to which a secant drawn through a point of tangency tends when the second point of intersection approaches the point of tangency indefinitely.

This property is taken as the definition of a tangent when talking about any curve.

Thus, the tangent to the curve AB (Fig.) is the limiting position MT to which the secant MN tends when the intersection point P approaches M without limit.

Note that the tangent defined in this way can have more than one common point with the curve (as can be seen in Fig.).

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

The common point is called in this case point of contact.

Possibility of existence tangent, and, moreover, drawn through any point circle, as a point of tangency, is proven as follows theorem.

Let it be required to carry out circle with center O tangent through the point A. To do this from the point A, as from the center, we describe arc radius A.O., and from the point O, as the center, we intersect this arc at the points B And WITH a compass solution equal to the diameter of the given circle.

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

Because O.D. And O.E.- radii, then D - middle O.B., A E- middle OS, Means AD And A.E. - medians, drawn to the bases of isosceles triangles, and therefore perpendicular to these bases. If straight D.A. And E.A. perpendicular to the radii O.D. And O.E., then they - tangents.

Consequence.

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

So AD=AE and ∠ OAD = ∠OAE because right triangles AOD And AOE, having a common hypotenuse A.O. and equal legs O.D. And O.E.(as radii), are equal. Note that here the word “tangent” actually means “ tangent segment” from a given point to the point of contact.

A straight line that has only one common point with a circle is called a tangent to the circle, and their common point is called the tangent point of the line and the circle.

Theorem (property of a tangent to a circle)

A tangent to a circle is perpendicular to the radius drawn to the point of tangency.

Given

A – point of contact

Prove:p OA

Proof.

Let's prove it by contradiction.

Let us assume that p is OA, then OA is inclined to the straight line p.

If from point O we draw a perpendicular OH to straight line p, then its length will be less than the radius: OH< ОА=r

We find that the distance from the center of the circle to the straight line p (OH) is less than the radius (r), which means the straight line p is secant (that is, it has two common points with the circle), which contradicts the conditions of the theorem (p is tangent).

This means that the assumption is incorrect, therefore the straight line p is perpendicular to OA.

Theorem (Property of tangent segments drawn from one point)

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

Given: approx. (O;r)

AB and AC are tangents to the surroundings. (O;r)

Prove: AB=AC

Proof

1) OB AB, OS AC, as radii drawn to the point of tangency (tangent property)

2) Consider tr. AOB, etc. AOS – p/u

JSC – general

OB=OS (as radii)

This means ABO = AOC (by hypotenuse and leg). Hence,

AB = AC,<3 = < 4 (как соответственные элементы в равных тр-ках). ч.т.д.

Theorem (Tangential test)

If a line passes through the end of a radius lying on a circle and is perpendicular to this radius, then it is a tangent.

Given: OA – radius of the circle

Prove: p- tangent to the circle

Proof

OA – radius of the circle (according to condition) (OA=r)

OA – perpendicular from O to straight line p (OA =d)

This means that r=OA=d, which means that the straight line p and the circle have one common point.

Therefore, line p is tangent to the circle. etc.

3.Properties of chords and secants.

Properties of tangent and secant

DEFINITION

Circumference is the locus of points equidistant from one point, which is called the center of the circle.

A line segment connecting two points on a circle is called chord(in the figure this is a segment). A chord passing through the center of a circle is called diameter circles.

1. The tangent is perpendicular to the radius drawn to the point of contact.

2. Tangent segments drawn from one point are equal.

3. 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.

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 2010 Unified State Exam, you can see that about 12% of the participants started the geometric problem C4, and only 0.2% of the participants received a full score, and in general the problem turned out to be the most difficult of all those proposed.

Obviously, the sooner we offer schoolchildren beautiful or unexpected ways of solving problems, the greater the likelihood of getting them interested and captivated seriously and for a long time. But how difficult it is to find interesting and complex problems 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 equality of triangles and the properties of adjacent and vertical angles? However, one can introduce the concept of a tangent to a circle, as a straight line that has one common point with the circle; assume that the radius drawn to the point of contact is perpendicular to the tangent. Of course, it is worth considering all possible cases of arrangement of two circles and common tangents to them, which can be drawn from zero to four. By proving the theorems proposed below, you can significantly expand the set of problems for seventh graders. At the same time, prove important or simply interesting and entertaining facts along the way. Moreover, since many statements are not included in the school textbook, they can be discussed in circle classes and with graduates when repeating planimetry. These facts turned out to be relevant last academic year. Since many diagnostic works and the work of the Unified State Examination itself contained a problem for the solution of which it was necessary to use the property of the tangent segment proved below.

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

This is the theorem that you can first introduce to seventh graders.
In the process of proof, we used the sign of equality of right triangles and concluded that the center of the circle lies on the bisector of the angle BSA.
Along the way, we remembered that the bisector of an angle is the locus of points in the interior region of the angle, equidistant from its sides. The solution to a far from trivial problem is based on these facts, accessible even to those just beginning to study geometry.

1. Angle bisectors A, IN And WITH 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 sums of the lengths of opposite sides are equal.

Solution (Fig. 2). Let ABOUT– point of intersection of these bisectors. Then ABOUT equidistant from all sides of the quadrilateral ABCD, that is
is the center of a circle inscribed in a quadrilateral. By theorem 1 the following equalities are true: AR = A.K., ER = E.P., F.T. = FK. Let's add the left and right sides term by term and get the correct equality:

(AR + ER) + F.T. = (A.K. +FK) + E.P.; A.E. + (F.C. + C.T.) = A.F. + (EU + PC). Because ST = RS, That AE + F.C. = A.F. + EU, which was what needed to be proven.

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-a triangle whose sides are sequentially 1, 2, 3, ..., n, into which a circle can be inscribed?

Solution. Let's say this n-gon exists. A 1 A 2 =1, …, A n-1 A n= n– 1,A n A 1 = n. B 1 , …, B n – corresponding points of contact. 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 A n= n – 1. Contradiction. Therefore no n-gon satisfying the conditions of the problem.

T 2 The sums of the opposite sides of a quadrilateral described 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. We can generalize this fact - the sums of the sides of a circumscribed even triangle, taken through one side, are equal. For example, for a hexagon ABCDEF right: AB + CD + EF = BC + DE + FA.

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

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

Solution (Fig. 5). By Theorem 1 MV = MX and RS = RH. Therefore, the perimeter of the triangle AMR equal to the sum of the segments AB And AC. Or double tangent drawn to the excircle for a triangle AMR . The value of the MOP angle is measured by half the angle VOS, which does not depend on the choice of point X.

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

Method two (geometric). Let's look at the diagram. Segments of equal tangents, taken one at a time, add up to the semi-perimeter
triangle. Red and green make up a side A. Then the segment we are interested in x = p – a. Of course, the results obtained coincide.

. Z

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

5. Prove that the points of tangency of the inscribed and excircle with the side of the triangle are symmetrical about the middle of this side.

Solution (Fig. 9). Note that AK is a tangent segment of the excircle for a triangle ABC. According to formula (2) . VM- line segment tangent to the incircle for a 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 external tangents and one internal tangent are drawn to two circles. The internal tangent intersects the external tangents at points A, B and touches the circles at points A 1 And IN 1 . Prove that AA 1 = BB 1.

Solution (Fig. 10). Stop... What is there to decide? This is just a different formulation of the previous problem. Obviously, one of the circles is inscribed and the other is excircle for a certain 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 Mathematics Olympiad for Schoolchildren is solved in such an obvious way.

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

Solution (Fig. 11). Suppose that in a pentagon ABCDE you can inscribe a circle. Moreover, the parties AB, B.C., CD, DE And EA are equal to 5, 6, 10, 7 and 8, respectively. Let us mark the tangent points in sequence – F, G, H, M And N. Let the length of the segment A.F. equal to X.

Then B.F. = FD – A.F. = 5 – x = B.G.. G.C. = B.C. – B.G. = = 6 – (5 – x) = 1 + x = CH. And so on: HD = DM = 9 – x; M.E. = EN = x – 2, AN = 10 – X.

But, A.F. = AN. That is 10 - X = X; X= 5. However, the tangent segment A.F. 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 order of circumambulation are 1, 2, 3, 4, 5. Find the length of the sixth side.

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

Then 1 + 3 + 5 = 2 + 4 + X, Where X– unknown sixth side, X = 3.

Solution (Fig. 12). Since the lengths of all sides are integers, the fractional parts of the lengths of the segments are equal BT, B.P., DM, DN, A.K. And AT. We have AT + TV= 1, and fractional parts of segment lengths AT And TB are equal. This is only possible when AT + TV= 0.5. By theorem 1 VT + 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 quadrangle ABCD AD = DC, AB = 3, BC = 5. Circles inscribed in triangles ABD And CBD touch a segment BD at points M And N respectively. Find the length of the segment MN.

Solution (Fig. 13). MN = DN – DM. According to formula (1) for triangles DBA And DBC accordingly, we have:

11. Into a quadrangle 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 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 up. The distance between the centers of the circles is equal to the sum of the radii. Answer: R+r.

In fact, it has been proven that the condition is in a quadrilateral ABCD you can inscribe a circle, 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. 14) 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 point marked D so that circles inscribed in triangles ABD And ACD touch a segment AD at one point. Find the length of the segment BD.

Solution (Fig. 15). Let's apply formula (1) for triangles ADC And A.D.B., 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 a triangle is connected to the tangency point of an inscribed circle on the opposite side, then the circles inscribed in the resulting triangles touch each other.

14. Centers ABOUT 1 , ABOUT 2 and ABOUT 3 three non-intersecting circles of the same radius are located at the vertices of a triangle. From points ABOUT 1 , ABOUT 2 , ABOUT 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 are painted 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 equal radii. Note that the segments BR And DM are equal, which follows from the equality of right triangles ABOUT 1 BR And O 2 B.M.. Likewise D.L. = D.P., FN = FK. We add the equalities term by term, then subtract from the resulting sums identical segments of tangents drawn from the vertices A, WITH, And E hexagon ABCDEF: AR And A.K., C.L. And C.M., EN And E.P.. We get what we need.

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

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

17. Unified State Exam. Diagnostic work 12/8/2009, P–4. Given a trapezoid ABCD, the foundations of which BC = 44,AD = 100, AB = CD= 35. Circle tangent to lines AD And A.C., touches the side CD at the point K. Find the length of the segment CK.BDC and BDA, touch the sides ВD at points E And F. Find the length of the segment E.F..

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

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

Problems for independent solution/

1. The perimeter of an isosceles trapezoid circumscribed about a circle is equal to 2 rub. Find the projection of the trapezoid's diagonal onto the larger base. (1/2r)

2. Open bank of Unified State Exam problems in mathematics. AT 4. To a circle inscribed in a triangle ABC (Fig. 22), three tangents are drawn. The perimeters of the cut triangles are 6, 8, 10. Find the perimeter of this triangle. (24)

3. Into a triangle ABC circle is inscribed. MN – tangent to the circle, MÎ AC, NÎ BC, BC = 13, AC = 14, AB = 15. Find the perimeter of the triangle MNC. (12)

4. To a circle inscribed in a square with side a, a tangent is drawn intersecting its two sides. Find the perimeter of the cut 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 tangency 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 long sides. Find the perimeter of the cut triangle. (16)

7. CD– 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 point marked D. To circles inscribed in triangles ABD And ACD, a common tangent is drawn intersecting AD at the point M. Find the length of the segment AM. (Length AM does not depend on the position of the point D And
equal to 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 extensions of the legs is equal to 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 touches a side AB at the point C 1. The excircle touches the extension of the side AB per point A at the point C 2. Determine the length of the segment C 1 C 2. (b)

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

12. Soros Olympiad 1996, 2nd round, 11th grade. Given a triangle 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, SA 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 the problem book by Gordin R.K. “Geometry. Planimetry." Moscow. Publishing house MCNMO. 2004.

The concept of a tangent to a circle

A circle has three possible relative positions relative to a straight line:

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

If the distance from the center of the circle to the straight line is equal to the radius, then the straight 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.

Let us now introduce the concept of a tangent line to a circle.

Definition 1

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

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

Figure 1. Tangent to a circle

Theorems related to the concept of a tangent to a circle

Theorem 1

Tangent property theorem: a tangent to a circle is perpendicular to the radius drawn to the point of tangency.

Proof.

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

Let us prove that $a\bot r$

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

Figure 2. Illustration of Theorem 1

That is, $OA$ is inclined to the tangent. Since the perpendicular to the straight line $a$ is always less than the inclined one to the same straight line, the distance from the center of the circle to the straight line is less than the radius. As we know, in this case the straight 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 of the tangent property theorem: If a 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 conditions of the problem, we have that the radius is a perpendicular drawn from the center of the circle to a given straight 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 find that this line is tangent to the circle.

The theorem has been proven.

Theorem 3

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

Proof.

Let a circle with center at point $O$ be given. Two different tangents are drawn from point $A$ (which lies on the entire circle). From the point of contact $B$ 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, then these triangles are equal in hypotenuse and leg.

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

The theorem has been proven.

Example of a problem on the concept of a tangent to a circle

Example 1

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

Solution.

Let's first depict everything in the figure (Fig. 4).

Figure 4.

Since $AC$ is a tangent and $OC$ is a radius, then by Theorem 1, we obtain that $\angle ACO=(90)^(()^\circ )$. We found that the triangle $ACO$ is rectangular, which means, by the Pythagorean theorem, we have:

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