Discussion Forum

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

Answer:

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 

 $(x-3)^{2}+(y+6)^{2}=60$

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