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Circles $C_{1}$ and $C_{2}$, of radii $r$ and $R$ respectively, touch each other as shown in figure. The line $\ell$, which is parallel to the line joining the centres of $C_{1}$ and $C_{2}$, is tangent to $\mathrm{C}_{1}$ at $\mathrm{P}$ and intersects $\mathrm{C}_{2}$ at $\mathrm{A}, \mathrm{B}$. If $\mathrm{R}^{2}=2 \mathrm{r}^{2}$, then $\angle A O B$
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The correct answer is:
$45^{\circ}$
$\mathrm{C}_{1}(\mathrm{r}, 0)$ $\mathrm{C}_{2}(\mathrm{R}, 0)$ $\mathrm{Eq} .$ of $\mathrm{AB} \quad \mathrm{y}=\mathrm{r}$
Eq. of circle $\quad C_{2} \quad(x-R)^{2}+y^{2}=R^{2}$
$\mathrm{A}\left(\mathrm{R}-\sqrt{\mathrm{R}^{2}-\mathrm{r}^{2}}, \mathrm{r}\right)$ using $\mathrm{R}^{2}=2 \mathrm{r}^{2}$
$\mathrm{B}\left(\mathrm{R}+\sqrt{\mathrm{R}^{2}-\mathrm{r}^{2}}, \mathrm{r}\right) \quad \mathrm{A}(\mathrm{R}-\mathrm{r}, \mathrm{r}), \mathrm{B}(\mathrm{R}+\mathrm{r}, \mathrm{r})$
Slope of $\mathrm{OA}=\frac{\mathrm{r}}{\mathrm{R}-\mathrm{r}}=\mathrm{m}_{1}$
Slope of $\mathrm{OB}=\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}=\mathrm{m}_{2}$
$$
\begin{array}{l}
\tan \theta=\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}=1 \\
\theta=45^{\circ}
\end{array}
$$
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