## Circles Exercise 10.6 – Questions:

I. Numerical problems on touching circles.

- 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.
- 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.
- 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.

- 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.
- 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.
- 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:**

**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.**

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

**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.**

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

**In 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˚

PC^{2} = OP^{2} – OC^{2}

x^{2} = (OQ – PQ)^{2} – (AC – OA)^{2}

[Since OP = OQ – PQ, OC = AC – AO]

x^{2} = (5 – x)^{2}+(6 – 5)^{2} [Since OQ = OA = 5]

x^{2} = 25 – 10x + x^{2} – 1

10x = 24

x = 2.4 cm

**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.**

Solution:

∠AOP = ∠BOQ [Vertically opposite angles]

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

∠BQO = ∠BOQ

∠APO = ∠BQO [alternate angles]

Therefore, AP||BQ

**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.**

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

**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**

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

**In 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˚

OC^{2} = OM^{2} + MC^{2} [By Pythagoras Theorem]

(CP + OP)^{2} = (MR – OR)^{2} + MC^{2}

(2 + x)^{2} = (4 – x)^{2} + 2^{2}

4 + 4x + x^{2} = 16 – 8x + x^{2} + 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]