Question

The equation of the circle concentric with the circle $x^{2}+y^{2}-6 x+12 y+15=0$ and of double its area is

• A. $x^{2}+y^{2} - 6 x+12 y - 15=0$
• B. $x^{2}+y^{2}-6 x+12 y-30=0$
• C. $x^{2}+y^{2}-6 x+12 y-25=0$
• D. $x^{2}+y^{2}-6 x+12 y-20=0$

Answer 1

Answer: (A)

The equation of the circle is
$S \equiv x^{2}+y^{2}-6 x+12 y+15=0$
Let the equation of concentric circle of given circles, is
$S_{2}=x^{2}+y^{2}-6 x+12 y+15=0$
On comparing the circle $S_{1}$ with,
$x^{2}+y^{2}+2 g x+2 f y+ c=0$
$\Rightarrow g=-3,\, f=6,\, c=15$
Then, radius of circle is
$=\sqrt{g^{2}+f^{2}-c}$
$=\sqrt{9+36-15}$
$=\sqrt{45-15}=\sqrt{30}$ units
and centre is $(-g,-f)=(3,-6)$
Now, the area of the circle $S$ is $=\pi$ (radius) $^{2}$
$=\pi(\sqrt{30})^{2}$
$=30\, \pi$
Let the radius of the concentric circle is $r_{2}$.
$r_{2} =\sqrt{g^{2}+f^{2}-c}$
$=\sqrt{9+36-c}$
$=\sqrt{45-c}$
Then, according to question, the area of concentric circle
$=2 \times$ area of $S$
$=2 \times 30\, \pi=60 \,\pi$
$\Rightarrow \pi r_{2}^{2}= 60 \,\pi$
$\Rightarrow (\sqrt{45-c})^{2}=60$
$\Rightarrow 45-c=60$
$\Rightarrow c=-15$
Hence, the equation of concentric circle is
$x^{2}+y^{2}+2(-3) x+2(6) y+(-15)=0$
$x^{2}+y^{2}-6 x+12 y-15=0$