Question

Find the equation of the circle concentric with the circle x² + y² − 6x + 12y + 15 = 0 and double of its area.

Answer 1

Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.
The centre of the circle x² + y² − 6x + 12y + 15 = 0 is (3, −6).

Area of the required circle = $2\mathrm{\pi }{r}^{\mathit{2}}$
Here, r = radius of the given circle

Now, r = $\sqrt{9+36-15}=\sqrt{30}$

∴ Area of the required circle = $2\mathrm{\pi }\left(30\right)=60\mathrm{\pi }$

Let R be the radius of the required circle.

∴ $60\mathrm{\pi }=\mathrm{\pi }{R}^{\mathit{2}}\mathit{\Rightarrow }{R}^{\mathit{2}}\mathit{=}60$

Thus, the equation of the required circle is ${\left(x-3\right)}^2+{\left(y+6\right)}^2=60$, i.e. $x^2+y^2-6x+12y=15$.