Answer:
Option A
Explanation:
Given,
$x^{2}+y^{2}-6x+12y+15=0$
centre (3,-6)
$r= \sqrt{9+36-15}= \sqrt{30}$
Area = $\pi (\sqrt{30})^{2}=30 \pi$
Required equation of circle has twice the area of circle
$\therefore$ $A'= 60 \pi$
$\pi R^{2}=60 \pi , R^{2}=60$
$\therefore$ Equation of circle concentric with given circle is
$(x-3)^{2}+(y+6)^{2}=60$
$x^{2}+y^{2}-6x+12y-15=0$
