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Let $A(1,2)$ be the centre and 3 be the radius of a circle $S$. Let $B(-1,-1)$ be the centre and $r$ be the radius of another circle $S^{\prime}$. If $\frac{\pi}{3}$ is the angle between the circles $S$ and $S^{\prime}$, then the number of possible values of $\mathrm{r}$ is
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Verified Answer
The correct answer is:
$2$
Let $C_1=A(1,2), r_1=3$
$\begin{aligned}
& C_2=B(-1,-1), r_2=r \\
& \Rightarrow \cos \theta=\frac{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2-r_1^2-r_2^2}{2 r_1 r_2} \\
& \Rightarrow \cos \left(\frac{\pi}{3}\right)=\frac{(-1-1)^2+(-2-1)^2-3^2-r^2}{2 \times 3 \times r}
\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{1}{2}=\frac{4-r^2}{6 r} \Rightarrow 2 r^2+6 r-8=0 \\ & \Rightarrow r^2+3 r-4=0 \Rightarrow(r-3)(r-1)=0 \\ & \Rightarrow \mathrm{r}=1,3\end{aligned}$
$\begin{aligned}
& C_2=B(-1,-1), r_2=r \\
& \Rightarrow \cos \theta=\frac{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2-r_1^2-r_2^2}{2 r_1 r_2} \\
& \Rightarrow \cos \left(\frac{\pi}{3}\right)=\frac{(-1-1)^2+(-2-1)^2-3^2-r^2}{2 \times 3 \times r}
\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{1}{2}=\frac{4-r^2}{6 r} \Rightarrow 2 r^2+6 r-8=0 \\ & \Rightarrow r^2+3 r-4=0 \Rightarrow(r-3)(r-1)=0 \\ & \Rightarrow \mathrm{r}=1,3\end{aligned}$
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