Circles Exercise 10.6 Solution – Chapter Circles – Class 10

Circles Exercise 10.6 – Questions:

I. Numerical problems on touching circles.

  1. Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.
  2. Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ∆ABC.
  3. In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.

II. Riders based on touching circles.

  1. A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
  2. Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.
  3. In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC

Circles Exercise 10.6 solution:

  1. Numerical problems on touching circles.

Circles Exercise 10.61.Three circles touch each other externally. Find the radii of the circles if the sides of the triangle formed by joining the centres are 7 cm, 8 cm and 9 cm respectively.

Solution:

Let radius of  the circles be AP = x ,  BQ = y and CR = z

AB = AP + BP = x + y = 7 cm

BC = BQ + CQ = y + z = 8 cm

AC = CR + AR = z + x = 9 cm

The perimeter of ∆ABC , AB + BC + CA = 7 + 8 + 9 = 24

AP + BP + BQ + CQ + CR + AR = 24

2x + 2y + 2z 24

x + y + z = 12

7 + z = 12⇒ z = 12 – 7 = 5 cm

x + 8 = 12 ⇒ x = 12 – 8 = 4 cm

y + 9 = 12 ⇒ y = 12 – 9 = 3 cm


  1. Circles Exercise 10.6Three circles with centres A, B and C touch each other as shown in the figure. IF the radii of these circles are 8 cm, 3 cm and 2 cm respectively, find the perimeter of ABC.

Solution:

The perimeter of ∆ABC = AB + BC + AC

AB = AM – BM = 8 – 3 = 5 cm

BC = BQ + CQ = 3 + 2 = 5 cm

AC = AN – CN = 8 – 2 = 6 cm

The perimeter of ∆ABC = AB + BC + AC = 5 + 5 + 6 = 16 cm


  1. Circles Exercise 10.6In the figure AB = 10 cm, AC = 6 cm and the radius of the smaller circle is x cm. Find x.

Solution:

In ∆OPC, ∠PCO = 90˚

PC2 = OP2 – OC2

x2 = (OQ – PQ)2 – (AC – OA)2

[Since OP = OQ – PQ, OC = AC – AO]

x2 = (5 – x)2+(6 – 5)2 [Since OQ = OA = 5]

x2 = 25 – 10x + x2 – 1

10x = 24

x = 2.4 cm


  1. Riders based on touching circles.

Circles Exercise 10.61.A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.

Solution:

∠AOP = ∠BOQ [Vertically opposite angles]

∠APO = ∠AOP [ AO = AP radius of the circle]

∠BQO = ∠BOQ

∠APO = ∠BQO [alternate angles]

Therefore, AP||BQ


  1. Circles Exercise 10.6Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.

Solution:

∠BXP = ∠PYC [Alternate angles AB||CD]

∠BPX = ∠PBX [ XB = XP radii]

∠BPX + ∠PBX + ∠BXP = 180˚ ………….(1)

∠CPY = ∠PCY [YP = YC radii]

∠CPY + ∠PCY + ∠PYC = 180˚

2∠CPY + ∠PCY = 180˚…………(2)

From (1) and (2),

2∠BPX + ∠BXP = 2∠CPY +  ∠PYC

2∠BPX = 2∠CPY

∠BPX = ∠CPY


  1. Circles Exercise 10.6In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2.OC

Solution:

∠ADB = 90˚

∠ACO = 90˚

In ∆ADB and ∆AOC,

∠ADB = ∠ACO = 90˚

∠A = ∠A

∆ADB ∼ ∆AOC

BD/OC = AB/AO

BD/OC = 2. AO/AO [AB = 2.OA]

BD/OC = 2

BD = 2.OC


  1. Circles Exercise 10.6In the given figure AB = 8 cm, M is the midpoint of AB. A circle with centre O touches all three semicircles as shown. Prove that the radius of this circle is shown. Prove that the radius of this circle is 1/6 AB

Solution:

In ∆OPC, ∠POC = 90˚

OC2 = OM2 + MC2 [By Pythagoras Theorem]

(CP + OP)2 = (MR – OR)2 + MC2

(2 + x)2 = (4 – x)2 + 22

4 + 4x + x2 = 16 – 8x + x2 + 4

4 + 4x = 16 – 8x + 4

12x = 16

x = 16/12  = 8/6

x = 1/6 x 8

x = 1/6 x AB [AB = 8 cm]