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
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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
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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˚
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
- 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
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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˚
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]
