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If a circle $\mathrm{C}$ passing through $(4,0)$ touches the circle $x^2+y^2+4 x-6 y-12=0$ externally at a point $(1,-1)$, then the radius of the circle $\mathrm{C}$ is :
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Verified Answer
The correct answer is:
5
5
Let A be the centre of given circle and $\mathrm{B}$ be the centre of circle C.
$$
\begin{aligned}
& x^2+y^2+4 x-6 y-12=0 \\
& \therefore A=(-2,3) \text { and } B=(g, f)
\end{aligned}
$$
Now, from the figure, we have
$$
\frac{-2+g}{2}=1 \text { and } \frac{3+f}{2}=-1
$$
(By mid point formula)
$$
\Rightarrow g=4 \text { and } f=-5
$$
Now, required radius
$$
=\mathrm{OB}=\sqrt{9+16}=\sqrt{25}=5
$$
$$
\begin{aligned}
& x^2+y^2+4 x-6 y-12=0 \\
& \therefore A=(-2,3) \text { and } B=(g, f)
\end{aligned}
$$
Now, from the figure, we have
$$
\frac{-2+g}{2}=1 \text { and } \frac{3+f}{2}=-1
$$
(By mid point formula)
$$
\Rightarrow g=4 \text { and } f=-5
$$
Now, required radius
$$
=\mathrm{OB}=\sqrt{9+16}=\sqrt{25}=5
$$
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