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If a diameter of the circle $x^2+y^2-4 x+6 y-12=0$ is a chord of a circle $S$ whose centre is at $(-3,2)$, then the radius of $S$ is
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The correct answer is:
$5 \sqrt {3}$
$r=\sqrt{(2)^2+(-3)^2+12}=\sqrt{25}=5$
$O C=\sqrt{(2+3)^2+(-3-2)^2}=5 \sqrt{2}$
$\begin{array}{ll}\therefore & O P^2=r^2+O C^2=25+50=75 \\ \Rightarrow & O P=\sqrt{75}=5 \sqrt{3} .\end{array}$
$O C=\sqrt{(2+3)^2+(-3-2)^2}=5 \sqrt{2}$
$\begin{array}{ll}\therefore & O P^2=r^2+O C^2=25+50=75 \\ \Rightarrow & O P=\sqrt{75}=5 \sqrt{3} .\end{array}$
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