EAMCET Engineering 2010 Solved Paper

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