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Parametric Equation of Parabola

Three circles...

Question

Three circles $C_{1}, C_{2}$ and $C_{3}$ with radii $r_{1}, r_{2}$ and $r_{1} + r_{2}$ respectively are such that $C_{1}$ and $C_{2}$ touch each other externally and $C_{3}$ internally. Another circle with radius $r_{3}$ touches all the three circles. If $r_{1} > r_{2} > r_{3}$ and $r_{1}, r_{2}, r_{3}$ are in A.P. then ${(\frac{r_{1}}{r_{2}})}^{3}$ is

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Solution

$cos (∠ C D A) = - cos (∠ C D B) \Rightarrow \frac{A D^{2} + D C^{2} - A C^{2}}{2 A D . D C} = - (\frac{B D^{2} + D C^{2} - B C^{2}}{2 B D . D C}) \Rightarrow \frac{(r_{1} + r_{2} - r_{1})^{2} + (r_{1} + r_{2} - r_{3})^{2} - (r_{1} + r_{3})^{2}}{2 r_{2} (r_{1} + r_{2} - r_{3})} = - \frac{(r_{1})^{2} + (r_{1} + r_{2} - r_{3})^{2} - (r_{2} + r_{3})^{2}}{2 r_{1} (r_{1} + r_{2} - r_{3})} 2 r_{2} = r_{1} + r_{3} as they are in A.P \Rightarrow \frac{(r_{2})^{2} + (2 r_{1} - r_{2})^{2} - (2 r_{2})^{2}}{2 r_{2} (2 r_{1} - r_{2})} = - \frac{(r_{1})^{2} + (2 r_{1} - r_{2})^{2} - (3 r_{2} - r_{1})^{2}}{2 r_{1} (2 r_{1} - r_{2})}$
By solving this we get
${(\frac{r_{1}}{r_{2}})}^{3} = 2$

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