Question

Find the equation of the circle with centre $(-a,-b)$ and radius $\sqrt{a^2-b^2}$.

Answer 1

The equation of a circle with centre $(h,k)$ and radius $r$ is given as
$(x-h{)}^2+(y-k{)}^2=r^2$
Here, centre $(h,k)=(-a,-b)$ and radius $\left(r\right)=\sqrt{a^2-b^2}$.
The equation of the circle is
$\Rightarrow (x+a{)}^2+(y+b{)}^2=(\sqrt{a^2-b^2}{)}^2$
$\Rightarrow x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$
$\Rightarrow x^2+y^2+2ax+2by+b^2=-b^2$
$\Rightarrow x^2+y^2+2ax+2by+2b^2=0$
$\therefore$ The equation of the required circle is $x^2+y^2+2ax+2by+2b^2=0$