Question

Find the equation of the circle cocentric with the circle $x^2+y^2-6x+12y+15=0$ and double of its area.

Answer 1

Equation of given circles

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

Hence coordinates sites

centre $(3,-6)$

Radius or circle $r=\sqrt{h^2+k^2-c}$

whose $h=3$, $k=-6$, $c=-15$

$\therefore r=\sqrt{9+(-6{)}^2-15}$

$\therefore r=\sqrt{30}$

Area of circle$=\pi r^2=30\pi$

It is given that the req circle is concentric to this circle and has double its area

$\therefore$ Area of required circle is $60\pi$

i.e., $\pi r^2=60\pi$

$\therefore r=\sqrt{4\times 15}=2\sqrt{15}$

Substituting g,f,, c, and r

$(2\sqrt{15}{)}^2=9+(-6{)}^2-c$

$\therefore c=-15$

since circle are concentric then values of $g,f$ are same

Hence equation required is $x^2+y^2-6x+12y-15=0$.