class 10 circles

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1. 10. CIRCLES

2. RECALL : What is a circle? A  circle  is a collection of all points in a plane which are at a constant distance called radius and from a certain fixed point called centre . A circumference which is a set of all points at a fixed distance from the centre of the circle. The distance from the centre of the circle to the circumference is called the radius of the circle. You can draw a circle using a compass. Circumference

3. APPLICATIONS OF CIRCLES IN DAY TO DAY LIFE :

4. IMPORTANT TERMS RELATED TO CIRCLE : Chord  of a circle is a line that joins two points on the circumference of a circle. The diameter of a circle is its longest chord. An  arc  of a circle is a continuous part of the circumference of the circle. A  sector  of a circle is the area/region between an arc and the center of the circle. A chord divides the area of a circle into two segments. The smaller area is called Minor segment and the bigger area is called Major segment.

5. Introduction to Circles Circle and line in a plane For a circle and a line on a plane, there can be  three  possibilities. i ) If the circle and line PQ have no point in common, then we say that PQ is a Non-intersecting line. ii) If the circle and line PQ have only one point in common, then we say that PQ is a Tangent to the circle. (iii) If the circle and line PQ have two distinct points A and B, then we say that PQ is a Secant of the circle . Also the line segment AB is called a chord of the circle.

6. TANGENT : A  tangent to a circle is a line which touches the circle at exactly one point. For every point on the circle, there is a unique tangent passing through it. A = Point of contact PQ = Tangent.

7. SECANT : SECANT : A  secant to a circle   is a line which has two distinct points in common with the circle. It cuts the circle at two points, forming a chord of the circle. PQ = Secant of the circle.

8. TWO PARALLEL TANGENTS AT MOST FOR A GIVEN SECANT : For every given  secant  of a circle, there are  exactly two tangents which are parallel  to it and touches the circle at two  diametrically opposite points .

9. IMPORTANT POINTS TO REMEMBER : The number of tangents drawn from a given point. i ) If the point is in an  interior region of the circle, any line through that point will be a secant . So,  no tangent  can be drawn to a circle which passes through a point that lies inside it. ii) When a point of tangency lies on the circle, there is  exactly one tangent  to a circle that passes through it.

10. TANGENT FROM AN EXTERNAL POINT When the point lies outside of the circle, there are accurately two tangents  to a circle through it Tangents to a circle from an external point

11. LENGTH OF A TANGENT The length of the tangent from the point (Say P) to the circle is defined as the segment of the tangent from the external point P to the point of tangency I with the circle. In this case, PI is the tangent length.

12. CHORD, TANGENT AND SECANT

13. CIRCLE :

14. OQ = OR + RQ = OP + RQ

15. Therefore OP ⊥ XY. Since we know that shortest distance from a point to a line is the perpendicular distance. Hence proved. This theorem is also called as TANGENT RADIUS THEOREM.

16. R H S This theorem is also called as EQUAL TANGENT LENGTHS THEOREM O O Q R P P

17. IMPORTANT POINTS TO REMEMBER : TANGENT RADIUS THEOREM : The tangent at any point of a circle is perpendicular to the radius through the point of contact . POINT OF CONTACT : The common point of a tangent to a circle and the circle is called point of contact. A line drawn through the end point of the radius and perpendicular to it is a tangent to the circle . EQUAL TANGENT LENGTHS THEOREM : The length of the tangents drawn from an external point are equal. One and only one tangent can be drawn at any point of a circle.

18. EX : 10.1 :   1.  How many tangents can a circle have ?    Answer : A circle can have infinitely many tangents since there are infinitely many points on a circle and at each point of it, it has a unique tangent. 2 .  Fill in the blanks : ( i ) A tangent to a circle intersects it in ONE point(s). (ii) A line intersecting a circle in two points is called a SECANT. (iii) A circle can have TWO parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called POINT OF CONTACT.

19. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm  (D) √119 cm Answer : Using the theorem, “The line drawn from the centre of the circle to the tangent is perpendicular to the tangent”. ∴ OP ⊥ PQ . By Pythagoras theorem in ΔOPQ, OQ 2  = OP 2  +   PQ 2 ⇒ (12) 2  = 5 2  + PQ 2 ⇒PQ 2  = 144 – 25 =  119 ⇒PQ = √119  cm (D) is the correct option.

20. 4. 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. Answer : AB and XY are two parallel lines where AB is the tangent to the circle at p oint C while XY is the secant to the circle.

21. EX : 10.2 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm Answer: First, draw a perpendicular from the center O of the triangle to a point P on the circle which is touching the tangent. This line will be perpendicular to the tangent of the circle. So, OP is perpendicular to PQ i.e. OP ⊥ PQ

22. By using Pythagorean theorem in △OPQ, OQ 2  = OP 2  + PQ 2 ⇒ (25) 2  = OP 2  + (24) 2 ⇒ OP 2  = 625 – 576 ⇒ OP 2  = 49 ⇒ OP = 7 cm So, option A i.e. 7 cm is the radius of the given circle. EX : 10.2 From the above figure, it is also seen that △OPQ is a right angled triangle. It is given that OQ = 25 cm and PQ = 24 cm

23. 2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90° EX : 10.2 Answer: From the question, it is clear that OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

24. So, OP ⊥ PT and TQ ⊥ OQ ∴ ∠ OPT = ∠ OQT = 90° Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360° So, ∠ PTQ + ∠ POQ + ∠ OPT + ∠ OQT = 360° Now, by putting the respective values we get, ⇒ ∠ PTQ + 90° + 110° + 90° = 360° ⇒ ∠ PTQ = 70° So, ∠ PTQ is 70° which is option B. EX : 10.2

25. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to (A) 50° (B) 60° (C) 70° (D) 80° Answer: GIVEN : EX : 10.2 OA is the radius to tangent PA OB is the radius to tangents PB. OA ⊥ PA and OB ⊥ PB ( Using Tangent Radius Theorem) So, ∠ OBP = ∠ OAP = 90° ∠ A BP = 80° TO FIND ∠ POA. In the quadrilateral AOBP, ∠ AOB + ∠ OAP + ∠ OBP + ∠ APB = 360° (Since the sum of all the interior angles will be 360°)

26. Putting their values we get, ⇒ ∠ AOB + 260° = 360° ⇒ ∠ AOB = 100° In △ OPB and △ OPA, AP = BP (Since the tangents from a point are always equal) OA = OB ( Equal radii ) OP = OP (common side) ∴ △ OPB ≅ △ OPA ( By SSS congruency). ∠ POB = ∠ POA (CPCT) ⇒ ∠ AOB = ∠ POA + ∠ POB ⇒ 2 ( ∠ POA) = ∠ AOB By putting the respective values we get, ⇒ ∠ POA = 100°/2 = 50° ∴ ∠ POA = 50° Option A is the correct option. EX : 10.2

27. Prove that the tangents drawn at the ends of a diameter of a circle are parallel . Answer: Radii of the circle to the tangents will be perpendicular to it. ∴ OB ⊥ RS and OA ⊥ PQ (By tangent radius theorem) ∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º From the figure, ∠OBR = ∠OAQ (Alternate interior angles) ∠OBS = ∠OAP (Alternate interior angles) Since alternate interior angles are equal, lines PQ and RS will be parallel. Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel. Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively. EX : 10.2

28. Q.5.    Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre . Sol.  GIVEN : Let the centre of the circle is O and tangent AB touches the circle at P. Let PX be perpendicular to AB at P. TOPROVE : PX passes through ‘O’ PROOF : If possible let PX not passing through O.         Join OP.            Since tangent at a point to a circle is perpendicular to the radius through that point,          ∴ AB ⊥ OP                i.e. ∠ OPB = 90°                ...(1)         But by construction,               AB ⊥ PX ⇒                  ∠ X PB = 90°                 ...(2)         From (1) and (2),                ∠ X PB = ∠ OPB         which is possible only when O and X coincide.         Thus, the perpendicular at the point of contact to the tangent passes through the centre. EX : 10.2

29. Q.6.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. Sol. ∵The tangent to a circle is perpendicular to the radius through the point of contact.               ∠OTA = 90°         Now, in the right ΔOTA, we have:               OP 2  = OT 2  + PT 2         ⇒ 5 2  = OT 2  + 4 2         ⇒ OT 2  = 5 2  – 4 2         ⇒ OT 2  = (5 – 4) (5 + 4)         ⇒ OT 2  = 1 × 9 = 9 = 3 2         ⇒ OT = 3         Thus, the radius of the circle is 3 cm. EX : 10.2

30. Q.7.    Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Sol.   In the figure, O is the common centre, of the given concentric circles.          AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.   Since OP is the radius of the smaller circle through P, the point of contact,  ∴ OP ⊥ AB (Tangent radius theorem)          ⇒ ∠OPB = 90° Also, a radius perpendicular to a chord bisects the chord. ∴ AP = BP, AB = AP + BP                    Now, in right ΔAPO,            OA 2  = AP 2  – OP 2         ⇒ 5 2  = AP 2  – 3 2         ⇒ AP 2  = 5 2  – 3 2         ⇒ AP 2  = (5 – 3) (5 + 3) = 2 × 8         ⇒ AP 2  = 16 = (4) 2         ⇒ AP = 4 cm                    Hence, the required length of the chord AB is 8 cm. EX : 10.2

31. I Q.8.    A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:   AB + CD = AD + BC Sol.  Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal. EX : 10.2 From the figure we observe that, (Using Equal tangent lengths theorem) DR = DS (Tangents on the circle from point D) … ( i ) AP = AS (Tangents on the circle from point A) … (ii) BP = BQ (Tangents on the circle from point B) … (iii) CR = CQ (Tangents on the circle from point C) … (iv) Adding all these equations, DR + AP + BP + CR = DS + AS + BQ + CQ ⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ) ⇒ AB + CD = AD + BC

32. In Fig. 10.13, XY and X′Y′ 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′Y′ at B. Prove that ∠ AOB = 90°.  Solution : EX : 10.2 A/q, In ΔOPA and ΔOCA, OP = OC (Radii of the same circle) AP = AC (Tangents from point A) AO = AO (Common side) ∴ ΔOPA ≅ ΔOCA (SSS congruence criterion) ⇒ ∠POA = ∠COA … ( i )

33. Similarly,  ΔOQB  ≅ ΔOCB ∠QOB = ∠COB … (ii) Since POQ is a diameter of the circle, it is a straight line. ∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º From equations ( i ) and (ii), 2∠COA + 2∠COB = 180º ⇒ ∠COA + ∠COB = 90º ⇒ ∠AOB = 90°

34. Given: Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively AB is the line segment, joining point of contacts A and B together such that it subtends ∠ AOB at center O of the circle. It can be observed that OA ⊥ PA ∴ ∠ OAP = 90° 10. 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: EX : 10.2

35. Similarly, OB ⊥ PB ∴ ∠ OBP = 90° In quadrilateral OAPB, Sum of all interior angles = 360º ∠ OAP + ∠ APB + ∠ PBO + ∠ BOA = 360º ⇒ 90º + ∠ APB + 90º + ∠ BOA = 360º ⇒ ∠ APB + ∠ BO A = 180º ∴ 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.

36. 11. Prove that the parallelogram circumscribing a circle is a rhombus. Answer: ABCD is a parallelogram, ∴ AB = CD ... ( i ) (opposite sides are equal) ∴ BC = AD ... (ii) From the figure, we observe that, (Using Equal tangent lengths theorem) DR = DS (Tangents to the circle at D) CR = CQ (Tangents to the circle at C) BP = BQ (Tangents to the circle at B) AP = AS (Tangents to the circle at A) EX : 10.2

37. Adding all these, DR + CR + BP + AP = DS + CQ + BQ + AS ⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) ⇒ CD + AB = AD + BC ... (iii) Putting the value of ( i ) and (ii) in equation (iii) we get, ⇒ 2AB = 2BC ⇒ AB = BC ... (iv) By Comparing equations ( i ), (ii), and (iv) we get, AB = BC = CD = DA ∴ ABCD is a rhombus.

38. 12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC. Answer In ΔABC, Length of two tangents drawn from the same point to the circle are equal, ∴ CF = CD = 6cm ∴ BE = BD = 8cm ∴ AE = AF =  x We observed that, AB = AE + EB =  x  + 8 BC = BD + DC = 8 + 6 = 14 CA = CF + FA = 6 +  x Now semi perimeter of triangle (s) is, ⇒ 2s = AB + BC + CA =  x  + 8 + 14 + 6 +  x = 28 + 2 x ⇒s = 14 +  x EX : 10.2

39. ⇒ also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB) = 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)] = 2×1/2 (4 x  + 24 + 32) = 56 + 4 x  ... (ii) Equating equation ( i ) and (ii) we get, √(14 +  x ) 48  x  = 56 + 4 x Squaring both sides, 48 x  (14 +  x ) = (56 + 4 x ) 2 ⇒ 48 x =  [4(14 + x)] 2 / (14 +  x ) ⇒ 48 x =  16 (14 +  x ) ⇒ 48 x =  224 + 16 x ⇒ 32 x =  224 ⇒  x =  7 cm Hence, AB =  x  + 8 = 7 + 8 = 15 cm CA = 6 +  x  = 6 + 7 = 13 cm Area of ΔABC = √s (s - a)(s - b)(s - c) = √(14 +  x ) (14 +  x  -   14)(14 +  x  -  x  - 6)(14 +  x  -  x -  8) = √(14 +  x ) ( x )(8)(6) = √(14 +  x ) 48  x  ... ( i )

40. 13.   Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.    Answer: Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the centre of the circle. In ΔOAP and ΔOAS, AP = AS (Tangents from the same point) OP = OS (Radii of the circle) OA = OA (Common side) ΔOAP ≅ ΔOAS (SSS congruence condition) ∴ ∠POA = ∠AOS ⇒∠1 = ∠8 EX : 10.2

41. Similarly we get, ∠2 = ∠3 ∠4 = ∠5 ∠6 = ∠7 Adding all these angles, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º ⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º ⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º ⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º ⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º ⇒ ∠AOB + ∠COD = 180º Similarly, we can prove that ∠ BOC + ∠ DOA = 180º Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.