Question

The equation of a circle concentric with the circle $x^{2}+y^{2}-6 x+12 y+15=0$ and having area that is twice the area of the given circle is

• A. $x^{2}+y^{2}-6 x+12 y-15=0$
• B. $x^{2}+y^{2}-6 x+12 y-30=0$
• C. $x^{2}+y^{2}-6 x+12 y-60=0$
• D. $x^{2}+y^{2}-6 x+12 y+15=0$

Answer 1

Answer: (A)

Given,
$x^{2}+y^{2}-6 x+12 y+15=0$
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 $
$x^{2}+y^{2}-6 x+12 y-15=0$