Question:
The sum of the areas of two circles which touch each other externally is \( 153\pi \). If the sum of their radii is 15, find the ratio of the larger to the smaller radius.
The sum of the areas of two circles which touch each other externally is \( 153\pi \). If the sum of their radii is 15, find the ratio of the larger to the smaller radius.
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Convert total area condition into an equation using identities. If radii sum is given, try substitution to reduce to a quadratic.
Updated On: Aug 26, 2025
- 4
- 2
- 3
- None of these
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The Correct Option is A
Solution and Explanation
Let the radii of the two circles be \( r_1 \) and \( r_2 \), with \( r_1>r_2 \).
Given: \[ r_1 + r_2 = 15 \quad \text{and} \quad \pi r_1^2 + \pi r_2^2 = 153\pi \Rightarrow r_1^2 + r_2^2 = 153 \] Let \( r_1 = x \), \( r_2 = 15 - x \). Then: \[ x^2 + (15 - x)^2 = 153
x^2 + 225 - 30x + x^2 = 153
2x^2 - 30x + 225 = 153
2x^2 - 30x + 72 = 0 \Rightarrow x^2 - 15x + 36 = 0 \] Solving the quadratic: \[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 36}}{2} = \frac{15 \pm \sqrt{225 - 144}}{2} = \frac{15 \pm \sqrt{81}}{2}
x = \frac{15 \pm 9}{2} \Rightarrow x = 12, \quad x = 3 \] So the radii are: \[ r_1 = 12, \quad r_2 = 3 \quad (\text{since } r_1>r_2) \Rightarrow \text{Ratio} = \frac{12}{3} = {4} \]
Given: \[ r_1 + r_2 = 15 \quad \text{and} \quad \pi r_1^2 + \pi r_2^2 = 153\pi \Rightarrow r_1^2 + r_2^2 = 153 \] Let \( r_1 = x \), \( r_2 = 15 - x \). Then: \[ x^2 + (15 - x)^2 = 153
x^2 + 225 - 30x + x^2 = 153
2x^2 - 30x + 225 = 153
2x^2 - 30x + 72 = 0 \Rightarrow x^2 - 15x + 36 = 0 \] Solving the quadratic: \[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 36}}{2} = \frac{15 \pm \sqrt{225 - 144}}{2} = \frac{15 \pm \sqrt{81}}{2}
x = \frac{15 \pm 9}{2} \Rightarrow x = 12, \quad x = 3 \] So the radii are: \[ r_1 = 12, \quad r_2 = 3 \quad (\text{since } r_1>r_2) \Rightarrow \text{Ratio} = \frac{12}{3} = {4} \]
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