Let circles $S_{1}$ and $S_{2}$ of radii $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively touches each other externally. Circle S of radius r touches $S_{1}$ and $S_{2}$ externally and also their direct common tangent. If the largest angle of triangle formed by joining centre of $S_{1}, S_{2}$ and S is A , then which of the following option(s) is / are true?

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A is greater than $90^{°}$

A can be less than $90^{°}$

$\tan \frac{A}{2} = \frac{4}{3}$ if $r_{1} = r_{2}$

$A = 90^{°}$ if $r_{1} = r_{2}$

answer is A, C.

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Detailed Solution

Since radii are in H.P.
$r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} - r^{2} - r_{1}^{2} - 2 {rr}_{1} - r^{2} - r_{2}^{2} - 2 {rr}_{2}$
$= - 2 {rr}_{1} - 2 {rr}_{2} + 2 r_{1} r_{2} - 2 r^{2} = 4 r \sqrt{r_{1} r_{2}} - 2 r^{2}$
$= 2 r (2 \sqrt{r_{1} r_{2}} - r) = positive \Rightarrow Δ C_{1} {CC}_{2}$ is obtuse angle triangle
If $r_{1} = r_{2}$ then from equation (1), $r_{1} = 4 r$ and $\sin \frac{A}{2} = \frac{r_{1}}{r + r_{1}}$