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If one of the diameter of the circle, given by the equation $x^{2}+y^{2}+4 x+6 y-12=0,$ is a chord of a circle $S,$ whose centre is $(2,-3$ ), the radius of $S$ is
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The correct answer is:
$\sqrt{41}$ unit
Given, equation of circle is $x^{2}+y^{2}+4 x+6 y-12=0$
Centre $C(-2,-3)$ and radius $=\sqrt{(-2)^{2}+(-3)^{2}+12}=5$
$$
x^{2}+y^{2}+4 x+6 y-12=0
$$
$$
\begin{aligned}
C_{1} C_{2} &=\sqrt{\left(2+{2})^{2}+(-3+3)^{2}\right.} \\
&=\sqrt{(4)^{2}+(0)^{2}}=4
\end{aligned}
$$
$\therefore$ Radius of circle, $S$
$$
=\sqrt{(4)^{2}+({5})^{2}}=\sqrt{16+25}=\sqrt{41} \text { unit }
$$
Centre $C(-2,-3)$ and radius $=\sqrt{(-2)^{2}+(-3)^{2}+12}=5$
$$
x^{2}+y^{2}+4 x+6 y-12=0
$$
$$
\begin{aligned}
C_{1} C_{2} &=\sqrt{\left(2+{2})^{2}+(-3+3)^{2}\right.} \\
&=\sqrt{(4)^{2}+(0)^{2}}=4
\end{aligned}
$$
$\therefore$ Radius of circle, $S$
$$
=\sqrt{(4)^{2}+({5})^{2}}=\sqrt{16+25}=\sqrt{41} \text { unit }
$$
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