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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

1. $4$
2. $2$
3. $3$
4. None of these

Let the radius of the first circle be $r_{1}$ and second circle be $r_{2}.$

Given that$:\pi r_{1}^{2} + \pi r_{2}^{2} = 153\pi$

$\implies r_{1}^{2} + r_{2}^{2} = 153\rightarrow(1)$

and $r_{1} + r_{2} = 1.5\rightarrow(2)$

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

$\implies (r_{1} + r_{2})^{2} = r_{1}^{2} + r_{2}^{2} + 2\:r_{1}r_{2}$

$\implies 15^{2 }= 153 + 2\:r_{1}r_{2}$

$\implies 2\:r_{1}r_{2} = 225 – 153$

$\implies 2\:r_{1}r_{2} = 72\rightarrow(3)$

We know that $(a – b)^{2} = a^{2} + b^{2} – 2ab$

$\implies (r_{1} – r_{2})^{2} = r_{1}^{2} + r_{2}^{2} – 2\:r_{1}r_{2}$

$\implies (r_{1} – r_{2})^{2} = 153 – 72 = 81$

$\implies r_{1} – r_{2} = 9\rightarrow(4)$

Now, solve the equation $(2)$ and $(4),$ we get $r_{1} = 12, r_{2} = 3$

The ratio of the larger to the smaller radius$:\dfrac{r_{1}}{r_{2}} = \dfrac{12}{3} = 4.$

So, the correct answer is $(A).$

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