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In the adjoining figure, two circles with centres $A$ and $B$ touch each other at the point $C$ . If the sum of their areas is $∴ {(R + r)}^{2} + {(R - r)}^{2} = 2 (R^{2} + r^{2})$ and the difference between their centres is $6 cm$ , then find the radii of two circles.

Accepted Answer

Given, two circles with centres $A$ and $B$ touch each other at the point $C$ .

Let the radius of the bigger circle be $R cm$ and the radius of the smaller circle be $6 cm$ .

Given, the sum of the areas of the circles is $116 π {cm}^{2}$ .

Area of a circle $= π r^{2}$ , where $r$ is the radius of the circle

$∴ π R^{2} + π r^{2} = 116 π$

$\Rightarrow π (R^{2} + r^{2}) = 116 π$

$\Rightarrow R^{2} + r^{2} = 116$ _____________ $(1)$

Given, the difference between their centres is $6 cm$ .

$∴ R - r = 6$ _____________ $(2)$

We know that ${(a + b)}^{2} + {(a - b)}^{2} = 2 (a^{2} + b^{2})$ .

$∴ {(R + r)}^{2} + {(R - r)}^{2} = 2 (R^{2} + r^{2})$

$\Rightarrow {(R + r)}^{2} + 6^{2} = 2 \times 116$ [From $(1)$ and $(2)$ ]

$\Rightarrow {(R + r)}^{2} = 232 - 36$

$\Rightarrow (R + r) = \sqrt{196}$

$\Rightarrow R + r = 14$ ________ $(3)$

Adding equation $(2)$ and equation $(3)$ , we get,

$R - r + R + r = 6 + 14$

$\Rightarrow 2 R = 20$

$\Rightarrow R = 10$

Substituting $R = 10$ in equation $(3)$ , we get,

$r + 10 = 14$

$\Rightarrow r = 14 - 10$

$\Rightarrow r = 4$

Hence, the radii of two circles are $10 cm$ and $4 cm$ respectively.

M L Aggarwal Solutions for Chapter: Areas Related to Circles, Exercise 6: Higher Order Thinking Skills (HOTS)

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Areas Related to Circles, Exercise 6: Higher Order Thinking Skills (HOTS)

Questions from M L Aggarwal Solutions for Chapter: Areas Related to Circles, Exercise 6: Higher Order Thinking Skills (HOTS) with Hints & Solutions

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