Circles

Welcome dear students! Today we are going to learn about Circles from Class ten Maths. You have studied in Class nine that a circle is a collection of all points in a plane which are at a constant distance, called the radius, from a fixed point, called the centre. You have also studied various terms related to a circle like chord, segment, sector, arc and so on. Let us now examine the different situations that can arise when a circle and a line are given in a plane. So, let us consider a circle and a line PQ. There can be three possibilities given in Figure 10.1. In Figure 10.1 (i), the line PQ and the circle have no common point. In this case, PQ is called a non-intersecting line with respect to the circle. In Figure 10.1 (ii), there are two common points A and B that the line PQ and the circle have. In this case, we call the line PQ a secant of the circle. In Figure 10.1 (iii), there is only one point A which is common to the line PQ and the circle. In this case, the line is called a tangent to the circle. You might have seen a pulley fitted over a well which is used in taking out water from the well. Look at Figure 10.2. Here the rope on both sides of the pulley, if considered as a ray, is like a tangent to the circle representing the pulley. Is there any position of the line with respect to the circle other than the types given above? You can see that there cannot be any other type of position of the line with respect to the circle. In this chapter, we will study about the existence of the tangents to a circle and also study some of their properties. [CHECKPOINT] In the previous section, you have seen that a tangent to a circle is a line that intersects the circle at only one point. To understand the existence of the tangent to a circle at a point, let us perform the following activities. Activity one: Take a circular wire and attach a straight wire AB at a point P of the circular wire so that it can rotate about the point P in a plane. Put the system on a table and gently rotate the wire AB about the point P to get different positions of the straight wire, as shown in Figure 10.3 (i). In various positions, the wire intersects the circular wire at P and at another point Q one or Q two or Q three. In one position, you will see that it will intersect the circle at the point P only, which is the position A prime B prime of AB. This shows that a tangent exists at the point P of the circle. On rotating further, you can observe that in all other positions of AB, it will intersect the circle at P and at another point, say R one or R two or R three. So, you can observe that there is only one tangent at a point of the circle. While doing the activity above, you must have observed that as the position AB moves towards the position A prime B prime, the common point, say Q one, of the line AB and the circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides with the point P in the position A prime B prime. Again note, what happens if AB is rotated rightwards about P? The common point R three gradually comes nearer and nearer to P and ultimately coincides with P. So, what we see is: The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide. Note that the word tangent comes from the Latin word tangere, which means to touch and was introduced by the Danish mathematician Thomas Fincke in the year fifteen eighty three. [CHECKPOINT] Activity two: On a paper, draw a circle and a secant PQ of the circle. Draw various lines parallel to the secant on both sides of it. You will find that after some steps, the length of the chord cut by the lines will gradually decrease, that is, the two points of intersection of the line and the circle are coming closer and closer, as shown in Figure 10.3 (ii). In one case, it becomes zero on one side of the secant and in another case, it becomes zero on the other side of the secant. See the positions P prime Q prime and P double prime Q double prime of the secant in Figure 10.3 (ii). These are the tangents to the circle parallel to the given secant PQ. This also helps you to see that there cannot be more than two tangents parallel to a given secant. This activity also establishes, what you must have observed while doing Activity one, namely, a tangent is the secant when both of the end points of the corresponding chord coincide. The common point of the tangent and the circle is called the point of contact, which is the point A in Figure 10.1 (iii), and the tangent is said to touch the circle at the common point. Now look around you. Have you seen a bicycle or a cart moving? Look at its wheels. All the spokes of a wheel are along its radii. Now note the position of the wheel with respect to its movement on the ground. Do you see any tangent anywhere? See Figure 10.4. In fact, the wheel moves along a line which is a tangent to the circle representing the wheel. Also, notice that in all positions, the radius through the point of contact with the ground appears to be at right angles to the tangent, as shown in Figure 10.4. We shall now prove this property of the tangent. [CHECKPOINT] Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Proof: We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY. Take a point Q on XY other than P and join OQ, as shown in Figure 10.5. The point Q must lie outside the circle. Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle. Therefore, OQ is longer than the radius OP of the circle. That is, OQ is greater than OP. Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY, as shown in Theorem A1.7. Remarks: One. By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent. Two. The line containing the radius through the point of contact is also sometimes called the normal to the circle at the point. [CHECKPOINT] Let us solve Exercise 10.1. Question one: How many tangents can a circle have? Answer: A circle can have infinitely many tangents, as there is exactly one tangent at each point on the circumference of the circle. Question two: Fill in the blanks. Part one: A tangent to a circle intersects it in one point. Part two: A line intersecting a circle in two points is called a secant. Part three: A circle can have two parallel tangents at the most. Part four: The common point of a tangent to a circle and the circle is called the point of contact. Question three: A tangent PQ at a point P of a circle of radius five centimetres meets a line through the centre O at a point Q so that OQ equals twelve centimetres. Length PQ is: Option A twelve centimetres, Option B thirteen centimetres, Option C eight point five centimetres, Option D square root of one hundred nineteen centimetres. Solution: Since PQ is tangent at P, OP is perpendicular to PQ. So triangle OPQ is a right angled triangle at P. Using Pythagoras theorem, OQ squared equals OP squared plus PQ squared. Substituting values, twelve squared equals five squared plus PQ squared. One hundred forty four equals twenty five plus PQ squared. PQ squared equals one hundred nineteen. So PQ equals square root of one hundred nineteen centimetres. The correct option is D. Question four: Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. Solution: Draw a circle with centre O. Draw a line L outside the circle. Draw a tangent to the circle parallel to L by drawing a radius perpendicular to L and extending it to touch the circle. Draw a secant parallel to L by drawing a line that cuts the circle at two distinct points and is parallel to L. [CHECKPOINT] Section 10.3: Number of Tangents from a Point on a Circle. To get an idea of the number of tangents from a point on a circle, let us perform the following activity. Activity three: Draw a circle on a paper. Take a point P inside it. Can you draw a tangent to the circle through this point? You will find that all the lines through this point intersect the circle in two points. So, it is not possible to draw any tangent to a circle through a point inside it, as shown in Figure 10.6 (i). Next take a point P on the circle and draw tangents through this point. You have already observed that there is only one tangent to the circle at such a point, as shown in Figure 10.6 (ii). Finally, take a point P outside the circle and try to draw tangents to the circle from this point. What do you observe? You will find that you can draw exactly two tangents to the circle through this point, as shown in Figure 10.6 (iii). We can summarise these facts as follows: Case one: There is no tangent to a circle passing through a point lying inside the circle. Case two: There is one and only one tangent to a circle passing through a point lying on the circle. Case three: There are exactly two tangents to a circle through a point lying outside the circle. In Figure 10.6 (iii), T one and T two are the points of contact of the tangents PT one and PT two respectively. The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle. Note that in Figure 10.6 (iii), PT one and PT two are the lengths of the tangents from P to the circle. The lengths PT one and PT two have a common property. Can you find this? Measure PT one and PT two. Are these equal? In fact, this is always so. Let us give a proof of this fact in the following theorem. [CHECKPOINT] Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. Proof: We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P, as shown in Figure 10.7. We are required to prove that PQ equals PR. For this, we join OP, OQ and OR. Then angle OQP and angle ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP, OQ equals OR as they are radii of the same circle. OP equals OP as it is common. Therefore, triangle OQP is congruent to triangle ORP by Right angle Hypotenuse Side congruence criterion. This gives PQ equals PR by Corresponding Parts of Congruent Triangles. Remarks: One. The theorem can also be proved by using the Pythagoras Theorem as follows: PQ squared equals OP squared minus OQ squared equals OP squared minus OR squared equals PR squared, since OQ equals OR, which gives PQ equals PR. Two. Note also that angle OPQ equals angle OPR. Therefore, OP is the angle bisector of angle QPR, that is, the centre lies on the bisector of the angle between the two tangents. [CHECKPOINT] Let us take some examples. Example one: Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. Solution: We are given two concentric circles C one and C two with centre O and a chord AB of the larger circle C one which touches the smaller circle C two at the point P, as shown in Figure 10.8. We need to prove that AP equals BP. Let us join OP. Then, AB is a tangent to C two at P and OP is its radius. Therefore, by Theorem 10.1, OP is perpendicular to AB. Now AB is a chord of the circle C one and OP is perpendicular to AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord, that is, AP equals BP. Example two: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ equals two times angle OPQ. Solution: We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P and Q are the points of contact, as shown in Figure 10.9. We need to prove that angle PTQ equals two times angle OPQ. Let angle PTQ equal theta. Now, by Theorem 10.2, TP equals TQ. So, triangle TPQ is an isosceles triangle. Therefore, angle TPQ equals angle TQP equals one half of one hundred eighty degrees minus theta, which equals ninety degrees minus theta over two. Also, by Theorem 10.1, angle OPT equals ninety degrees. So, angle OPQ equals angle OPT minus angle TPQ equals ninety degrees minus ninety degrees minus theta over two, which equals theta over two, that is, one half of angle PTQ. This gives angle PTQ equals two times angle OPQ. [CHECKPOINT] Example three: PQ is a chord of length eight centimetres of a circle of radius five centimetres. The tangents at P and Q intersect at a point T, as shown in Figure 10.10. Find the length TP. Solution: Join OT. Let it intersect PQ at the point R. Then triangle TPQ is isosceles and TO is the angle bisector of angle PTQ. So, OT is perpendicular to PQ and therefore, OT bisects PQ which gives PR equals RQ equals four centimetres. Also, OR equals square root of OP squared minus PR squared equals square root of five squared minus four squared equals three centimetres. Now, angle TPR plus angle RPO equals ninety degrees equals angle TPR plus angle PTR. Why? Because in right triangle OPT, the acute angles sum to ninety degrees. So, angle RPO equals angle PTR. Therefore, right triangle TRP is similar to the right triangle PRO by Angle Angle similarity. This gives TP divided by PO equals RP divided by RO, that is, TP divided by five equals four divided by three, or TP equals twenty over three centimetres. Note: TP can also be found by using the Pythagoras Theorem, as follows: Let TP equal x and TR equal y. Then x squared equals y squared plus sixteen, taking right triangle PRT. And x squared plus five squared equals y plus three squared, taking right triangle OPT. Subtracting the first equation from the second, we get twenty five equals six y minus seven, or y equals thirty two over six which simplifies to sixteen over three. Therefore, x squared equals sixteen over three squared plus sixteen, which equals two hundred fifty six over nine plus one hundred forty four over nine, which equals four hundred over nine. So x equals twenty over three. [CHECKPOINT] Let us solve Exercise 10.2. In questions one to three, choose the correct option and give justification. Question one: From a point Q, the length of the tangent to a circle is twenty four centimetres and the distance of Q from the centre is twenty five centimetres. The radius of the circle is: Option A seven centimetres, Option B twelve centimetres, Option C fifteen centimetres, Option D twenty four point five centimetres. Solution: Let the point of contact be P. Then OP is perpendicular to PQ. In right triangle OPQ, OQ squared equals OP squared plus PQ squared. Twenty five squared equals r squared plus twenty four squared. Six hundred twenty five equals r squared plus five hundred seventy six. r squared equals forty nine. r equals seven centimetres. Correct option is A. Question two: In Figure 10.11, if TP and TQ are the two tangents to a circle with centre O so that angle POQ equals one hundred ten degrees, then angle PTQ is equal to: Option A sixty degrees, Option B seventy degrees, Option C eighty degrees, Option D ninety degrees. Solution: In quadrilateral OPTQ, angle OPT and angle OQT are ninety degrees each. Sum of angles in a quadrilateral is three hundred sixty degrees. So angle PTQ plus angle POQ plus ninety plus ninety equals three hundred sixty. angle PTQ plus one hundred ten plus one hundred eighty equals three hundred sixty. angle PTQ equals seventy degrees. Correct option is B. Question three: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of eighty degrees, then angle POA is equal to: Option A fifty degrees, Option B sixty degrees, Option C seventy degrees, Option D eighty degrees. Solution: In quadrilateral OAPB, angle OAP and angle OBP are ninety degrees. angle APB is eighty degrees. So angle AOB equals three hundred sixty minus ninety minus ninety minus eighty equals one hundred degrees. Triangles OAP and OBP are congruent, so OP bisects angle AOB. Therefore, angle POA equals half of one hundred degrees, which is fifty degrees. Correct option is A. [CHECKPOINT] Question four: Prove that the tangents drawn at the ends of a diameter of a circle are parallel. Solution: Let AB be a diameter of the circle with centre O. Let tangents at A and B be lines l and m respectively. By Theorem 10.1, the tangent at A is perpendicular to radius OA, so l is perpendicular to AB. Similarly, the tangent at B is perpendicular to radius OB, so m is perpendicular to AB. Since both l and m are perpendicular to the same line AB, they are parallel to each other. Hence proved. Question five: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. Solution: Let the tangent be XY touching the circle at P. Let the perpendicular to XY at P be line l. By Theorem 10.1, the radius OP is perpendicular to XY at P. Since only one perpendicular can be drawn to a line at a given point on it, line l must coincide with OP. Therefore, the perpendicular at the point of contact passes through the centre O. Hence proved. Question six: The length of a tangent from a point A at distance five centimetres from the centre of the circle is four centimetres. Find the radius of the circle. Solution: Let the point of contact be B. Then OB is perpendicular to AB. In right triangle OBA, OA squared equals OB squared plus AB squared. Five squared equals r squared plus four squared. Twenty five equals r squared plus sixteen. r squared equals nine. r equals three centimetres. Question seven: Two concentric circles are of radii five centimetres and three centimetres. Find the length of the chord of the larger circle which touches the smaller circle. Solution: Let the chord of the larger circle be AB touching the smaller circle at P. Join OP. OP is perpendicular to AB. In right triangle OPA, OA equals five centimetres, OP equals three centimetres. AP squared equals OA squared minus OP squared equals twenty five minus nine equals sixteen. So AP equals four centimetres. Since perpendicular from centre bisects the chord, AB equals two times AP equals eight centimetres. [CHECKPOINT] Question eight: A quadrilateral ABCD is drawn to circumscribe a circle, as shown in Figure 10.12. Prove that AB plus CD equals AD plus BC. Solution: Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively. By Theorem 10.2, lengths of tangents from external points are equal. So AP equals AS, BP equals BQ, CQ equals CR, DR equals DS. Adding these, AP plus BP plus CR plus DR equals AS plus BQ plus CQ plus DS. That is, AB plus CD equals AD plus BC. Hence proved. Question nine: In Figure 10.13, XY and X prime Y prime are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X prime Y prime at B. Prove that angle AOB equals ninety degrees. Solution: Tangents from A to the circle are AP and AC, so angle AOP equals angle AOC. Tangents from B to the circle are BQ and BC, so angle BOQ equals angle BOC. Since XY and X prime Y prime are parallel, angle POQ is a straight line, so one hundred eighty degrees. Angle POQ equals angle AOP plus angle AOC plus angle BOC plus angle BOQ. Substituting, two angle AOC plus two angle BOC equals one hundred eighty. So two times angle AOB equals one hundred eighty. Therefore, angle AOB equals ninety degrees. Question ten: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. Solution: Let tangents from P touch circle at A and B. Join OA, OB, OP. In quadrilateral OAPB, angle OAP and angle OBP are ninety degrees. Sum of angles is three hundred sixty degrees. So angle APB plus angle AOB plus ninety plus ninety equals three hundred sixty. angle APB plus angle AOB equals one hundred eighty degrees. Hence they are supplementary. [CHECKPOINT] Question eleven: Prove that the parallelogram circumscribing a circle is a rhombus. Solution: Let parallelogram ABCD circumscribe a circle touching sides at P, Q, R, S. By tangent property, AP equals AS, BP equals BQ, CQ equals CR, DR equals DS. Adding, AP plus BP plus CR plus DR equals AS plus BQ plus CQ plus DS. So AB plus CD equals AD plus BC. In a parallelogram, opposite sides are equal: AB equals CD and AD equals BC. So two AB equals two AD, which means AB equals AD. Since adjacent sides are equal, ABCD is a rhombus. Question twelve: A triangle ABC is drawn to circumscribe a circle of radius four centimetres such that the segments BD and DC into which BC is divided by the point of contact D are of lengths eight centimetres and six centimetres respectively, as shown in Figure 10.14. Find the sides AB and AC. Solution: Let the circle touch AB at E and AC at F. By tangent property, BD equals BE equals eight centimetres, CD equals CF equals six centimetres. Let AE equals AF equals x. Then AB equals x plus eight, AC equals x plus six, BC equals fourteen. Area of triangle ABC equals sum of areas of OAB, OBC, OCA. Area equals half times perimeter times inradius. Semi perimeter s equals x plus eight plus x plus six plus fourteen divided by two equals x plus fourteen. Area equals four times x plus fourteen. Also, area squared equals s times s minus a times s minus b times s minus c. s minus a equals x, s minus b equals eight, s minus c equals six. So area squared equals x plus fourteen times x times eight times six. So sixteen times x plus fourteen squared equals forty eight x times x plus fourteen. Divide by x plus fourteen: sixteen times x plus fourteen equals forty eight x. sixteen x plus two hundred twenty four equals forty eight x. thirty two x equals two hundred twenty four. x equals seven. So AB equals seven plus eight equals fifteen centimetres. AC equals seven plus six equals thirteen centimetres. [CHECKPOINT] Question thirteen: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Solution: Let quadrilateral ABCD circumscribe a circle touching sides at P, Q, R, S. Join OA, OB, OC, OD, OP, OQ, OR, OS. By tangent property, triangles OAP and OAS are congruent, so angle AOP equals angle AOS. Similarly, angle BOP equals angle BOQ, angle COQ equals angle COR, angle DOR equals angle DOS. Sum of all angles at O is three hundred sixty degrees. So two angle AOP plus two angle BOP plus two angle COQ plus two angle DOR equals three hundred sixty. Dividing by two, angle AOB plus angle COD equals one hundred eighty degrees. Similarly, angle BOC plus angle DOA equals one hundred eighty degrees. Hence opposite sides subtend supplementary angles at the centre. Section 10.4 Summary: In this chapter, you have studied the following points: One. The meaning of a tangent to a circle. Two. The tangent to a circle is perpendicular to the radius through the point of contact. Three. The lengths of the two tangents from an external point to a circle are equal. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]