Chapter 3 Circle Practice Set 3.2 Solutions

Let the two circles having centres P and Q touch each other internally at point R. Here, QR = 3.5 cm, PR = 4.8 cm The two circles touch each other internally. ∴ By theorem of touching circles, P – Q – R PQ = PR – QR = 4.8 – 3.5 = 1.3 cm [The distance between the centres of circles touching internally is equal to the difference in their radii]

Let the two circles having centres P and R touch each other externally at point Q. Here, PQ = 5.5 cm, QR = 4.2 cm The two circles touch each other externally. ∴ By theorem of touching circles, P – Q – R PR = PQ + QR = 5.5 + 4.2 = 9.7 cm [The distance between the centres of the circles touching externally is equal to the sum of their radii]

i. Circles touching externally: ii. Circles touching internally:

The circles with centres P and Q touch each other at R. ∴ By theorem of touching circles, P – R – Q i. In ∆PAR, seg PA = seg PR [Radii of the same circle] ∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem] Similarly, in ∆QBR, seg QR = seg QB [Radii of the same circle] ∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem] But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles] ∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)] But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ. ∴ seg AP || seg BQ [Alternate angles test] ii. In ∆APR and ∆RQB, ∠PAR ≅ ∠QRB [From (i) and (iii)] ∠APR ≅ ∠RQB [Alternate angles] ∴ ∆APR – ∆RQB [AA test of similarity] iii. ∠PAR = 35° [Given] ∴ ∠RBQ = ∠PAR= 35° [From (iv)] In ∆RQB, ∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°] ∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)] ∴ ∠RQB + 2 ∠RBQ = 180° ∴ ∠RQB + 2 × 35° = 180° ∴ ∠RQB + 70° = 180° ∴ ∠RQB = 110°

i. The circles with centres A and B touch each other at E. [Given] ∴ By theorem of touching circles, A – E – B ∴ ∠ACD = ∠BDC = 90° [Tangent theorem] ∠AFD = 90° [Construction] ∴ ∠CAF = 90° [Remaining angle of ꠸AFDC] ∴ ꠸AFDC is a rectangle. [Each angle is of measure 900] ∴ AC = DF = 4 cm [Opposite sides of a rectangle] Now, BD = BF + DF [B – F – C] ∴ 6 = BF + 4 BF = 2 cm Also, AB = AE + EB = 4 + 6 = 10 cm [The distance between the centres of circles touching externally is equal to the sum of their radii]

Line l is a common tangent of the two circles.

Line l is a common tangent of the two circles.

Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R – T – S]. Hence the given circles are externally touching circles.

Circles with centres N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K – N – M]. Hence, the given circles are internally touching circles.

i. Here, circle with centres A and B touch each other externally at point C. ∴ d(A, B) = d(A, C) + d(B,C) = 3 + 4 ∴ d(A,B) = 7 cm [The distance between the centres of circles touching externally is equal to the sum of their radii] ii. Here, circle with centres A and 13 touch each other internally at point C. ∴ d(A, B) = d(A, C) – d(B, C) = 4 – 3 ∴ d(A,B) = 1 cm [The distance between the centres of circles touching internally is equal to the difference in their radii]