TOPICS
Exercise - 10.2

Question-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

Solution :-

  Let O be the centre of the circle. 
  Given that : 
  OQ = 25cm and PQ = 24 cm 
  As the radius is perpendicular to the tangent at the point of contact, 
  Therefore, OP ⊥ PQ 

  Applying Pythagoras theorem in ∆OPQ, we obtain
   OP = √PQ² - OQ² 
      = √25² - 24² 
      = √625 - 576 
      = √49
   OP = 7 cm
  Therefore, the radius of the circle is 7 cm. 
  Hence, Option (A) is correct Answer. 
    

Question-2 :- In Figure, 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°

Solution :-

  Given that : 
  TP and TQ are tangents. 
  Therefore, radius drawn to these tangents will be perpendicular to the tangents. 
  Thus, OP ⊥ TP and OQ ⊥ TQ 
  ∠OPT = 90°, ∠OQT = 90°, ∠ POQ = 110°

  In quadrilateral POQT, 
  Sum of all interior angles = 360° 
  ∠OPT + ∠POQ + ∠OQT + ∠PTQ = 360°
  90° + 110° + 90° + ∠PTQ = 360° 
  290° + ∠PTQ = 360°
  ∠PTQ = 360° - 290°
  ∠PTQ = 70°
  Hence, Option (B) is correct Answer
    

Question-3 :-  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°

Solution :-

  Given that : 
  PA and PB are tangents. 
  Therefore, the radius drawn to these tangents will be perpendicular to the tangents. 
  Thus, OA ⊥ PA and OB ⊥ PB 
  ∠OBP = 90°, ∠OAP = 90°, APB = 80° 

  In Quadrilateral AOBP, 
  Sum of all interior angles = 360° 
  ∠OAP + ∠APB +∠PBO + ∠BOA = 360° 
  90° + 80°	+ 90° + BOA = 360° 
  260° + ∠BOA = 360°
  ∠BOA = 360° - 260°
  ∠BOA = 100° 

  In ∆OPB and ∆OPA, 
  AP = BP (Tangents from a point) 
  OA = OB (Radii of the circle) 
  OP = OP (Common side) 
  Therefore, ∆OPB ≅ ∆OPA (SSS congruence criterion) 
  And thus, ∠POB = ∠POA 
  ∠POA = 1/2 ∠BOA = 100°/2 = 50°
  Hence, Option (A) is correct Answer. 
    

Question-4 :-  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. 
  Two tangents PQ and RS are drawn at points A and B respectively. 
  Radius drawn to these tangents will be perpendicular to the tangents. 
  Thus, OA ⊥ RS and OB ⊥ PQ 
  ∠OAR = 90°, ∠OAS = 90°, ∠OBP = 90°, ∠OBQ = 90°
  It can be observed that 
  ∠OAR = ∠OBQ (Alternate interior angles) 
  ∠OAS = ∠OBP (Alternate interior angles) 
  Since alternate interior angles are equal, lines PQ and RS will be parallel. 
  i.e., PQ || RS
    

Question-5 :- Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution :-

  Let us consider a circle with centre O. 
  Let AB be a tangent which touches the circle at P. 
  Let us assume that the perpendicular to AB at P does not pass through centre O.
  Let it pass through another point O'. 
  Join OP and O'P. 

  As perpendicular to AB at P passes through O', 
  Therefore, ∠O'PB = 90° ..... (1) 
  O is the centre of the circle and P is the point of contact. 

  We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other. 
  ∴ ∠OPB = 90° ..... (2) 

  Comparing equations (1) and (2), we obtain 
  ∠O'PB = ∠OPB ..... (3) 

  From the figure, it can be observed that, 
  ∠O'PB < ∠OPB ..... (4) 

  Therefore, ∠O'PB = ∠OPB is not possible. 
  It is only possible, when the line O'P coincides with OP. 
  Therefore, the perpendicular to AB through P passes through centre O. 
    

Question-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.

Solution :-

  Given that : 
  OA = 5cm and AB = 4 cm 
  Let us consider a circle centered at point O. 
  AB is a tangent drawn on this circle from point A. 

  In ∆ ABO, 
  OB ⊥ AB (Radius ⊥ tangent at the point of contact) 
  Applying Pythagoras theorem in ∆ABO, we obtain 
  OB = √OA² - AB² 
     = √5² - 4² 
     = √25 - 16 
     = √9
  OB = 3 cm
  Hence, the radius of the circle is 3 cm. 
    

Question-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.

Solution :-

  Let the two concentric circles be centered at point O. 
  Let PQ be the chord of the larger circle which touches the smaller circle at point A. 
  Therefore, PQ is tangent to the smaller circle. 
  OA ⊥ PQ (As OA is the radius of the circle
         
  Applying Pythagoras theorem in ∆OAP, we obtain 
  AP = √OP² - OA² 
     = √5² - 3² 
     = √25 - 9 
     = √16
  AP = 4 cm

  In ∆OPQ, 
  Since OA ⊥ PQ, 
  AP = AQ (Perpendicular from the center of the circle bisects the chord) 
  PQ = 2AP = 2 × 4 = 8 
  Therefore, the length of the chord of the larger circle is 8 cm. 
    

Question-8 :- A quadrilateral ABCD is drawn to circumscribe a circle (see Figure). Prove that AB + CD = AD + BC

Solution :-

  It can be observed that 
  DR = DS (Tangents on the circle from point D) .... (1) 
  CR = CQ (Tangents on the circle from point C) .... (2) 
  BP = BQ (Tangents on the circle from point B) .... (3) 
  AP = AS (Tangents on the circle from point A) .... (4) 

  Adding all these equations, we obtain 
  DR + CR + BP + AP = DS + CQ + BQ + AS 
  (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) 
  CD + AB = AD + BC 
    

Question-9 :-  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 :-

  In ∆OPA and ∆OCA, 
  OP = OC (Radii of the same circle) 
  AP = AC (Tangents from point A) 
  AO = AO (Common side) 
  ∆ OPA ≅ ∆ OCA (SSS congruence criterion) 

  ∠POA = ∠COA ..... (i ) 
  Similarly, ∆ OQB ≅ ∆ OCB 
  ∠QOB = ∠COB ....... (ii) 
  Since POQ is a diameter of the circle, it is a straight line. 
  Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
  From equations (i) and (ii), it can be observed that 
  2∠COA + 2∠COB = 180°
  ∠COA + ∠COB = 90°
  ∠AOB = 90° 
    

Question-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 :-

  Let us consider a circle centered at point 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 and 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 (radius) ⊥ PA (tangent) 
  Therefore, ∠OAP = 90° 
  Similarly, OB (radius) ⊥ PB (tangent) 
  ∠OBP = 90° 

  In quadrilateral OAPB, 
  Sum of all interior angles = 360°
  ∠OAP + ∠APB + ∠PBO + ∠BOA = 360° 
  90° + ∠APB + 90° + ∠BOA = 360°
  ∠APB + ∠BOA = 180°
  Hence, it can be observed 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. 
    

Question-11 :-  Prove that the parallelogram circumscribing a circle is a rhombus.

Solution :-

  Here, ABCD is a parallelogram, 
  so, AB = CD ....(1) 
  and BC = AD ....(2) 

  It can be observed that 
  DR = DS (Tangents on the circle from point D) 
  CR = CQ (Tangents on the circle from point C) 
  BP = BQ (Tangents on the circle from point B) 
  AP = AS (Tangents on the circle from point A) 

  Adding all these equations, we obtain 
  DR + CR + BP + AP = DS + CQ + BQ + AS  
  (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) 
  CD + AB = AD + BC 
  On putting the values of equations (1) and (2) in this equation, we obtain 
  2AB = 2BC 
  AB = BC ....(3) 
  Comparing equations (1), (2), and (3), we obtain 
  AB = BC = CD = DA 
  Hence, ABCD is a rhombus. 
    

Question-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 Figure). Find the sides AB and AC.

Solution :-

  Let the given circle touch the sides AB and AC of the triangle at point E and F 
  respectively and the length of the line segment AF be x. 

  In ∆ ABC, 
  CF = CD = 6cm (Tangents on the circle from point C)
  BE = BD = 8cm (Tangents on the circle from point B) 
  AE = AF = x (Tangents on the circle from point A) 

  AB = AE + EB = x + 8, 
  i.e., c = x + 8

  BC = BD + DC = 8 + 6 = 14,
  i.e., a = 14

  CA = CF + FA = 6 + x, 
  i.e., b = 6 + x 

  2s = AB + BC + CA 
     = x + 8 + 14 + 6 + x 
     = 28 + 2x 
   s = 14 + x  

  Area of ∆ ABC = √s(s - a)(s - b)(s - c) 
                = √(14 + x) {(14 + x) - 14} {(14 + x) - (x + 6)} {(14 + x) - (x + 8)} 
                = √(14 + x) {14 + x - 14} {14 + x - x - 6)} {14 + x - x - 8} 
                = √48x (14 + x) 
                = 4√3(14x + x²)

  Area of ∆ OBC = 1/2 x OD x BC = 1/2 x 4 x 14 = 28
  Area of ∆ OCA = 1/2 x OF x AC = 1/2 x 4 x (6 + x) = 12 + 2x
  Area of ∆ OBA = 1/2 x OE x AB = 1/2 x 4 x (8 + x) = 16 + 2x

  Area of ∆ ABC = Area of ∆ OBC + Area of ∆ OCA + Area of ∆ OAB 
  4√3(14x + x²) = 28 + 12 + 2x + 16 + 2x
  4√3(14x + x²) = 56 + 4x
  √3(14x + x²) = 14 + x
  3(14x + x²) = (14 + x)²
  42x + 3x² = 196 + x² + 28x
  3x² - x² + 42x - 28x - 196 = 0
  2x² + 14x - 196 = 0
  x² + 7x - 98 = 0
  x² + 14x - 7x - 98 = 0
  x(x + 14) - 7(x + 14) = 0
  (x + 14)(x - 7) = 0
  x + 14 = 0 or x - 7 = 0
  x = -14 or x = 7 

  However, x = -14 is not possible as the length of the sides will be negative. 
  Therefore, x = 7 
  Hence, AB = x + 8 = 7 + 8 = 15 cm 
  CA = 6 + x = 6 + 7 = 13 cm 
    

Question-13 :-  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution :-

  Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. 
  Let us join the vertices of the quadrilateral ABCD to the center of the circle. 

  Consider ∆OAP and ∆OAS, 
  AP = AS (Tangents from the same point) 
  OP = OS (Radii of the same circle) 
  OA = OA (Common side) 
  ∆OAP ≅ ∆OAS (SSS congruence criterion) 

  And thus, ∠POA = ∠AOS 
  ∠1 = ∠8 
  Similarly, 
  ∠2 = ∠3 
  ∠4 = ∠5 
  ∠6 = ∠7 
  Now,
  ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
  (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360° 
  (∠1 + ∠1) + (∠2 + ∠2) + (∠5 + ∠5) + (∠6 + ∠6) = 360° 
  2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360° 
  2(∠1 + ∠2) + 2(∠5 + ∠6) = 360° 
  (∠1 + ∠2) + (∠5 + ∠6) = 180° 
  ∠AOB + ∠COD = 180°
  Similarly, we can prove that 	∠BOC + ∠DOA = 180° 
  Hence, opposite sides of a quadrilateral circumscribing a circle subtend 
  supplementary angles at the centre of the circle.

    
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