Question

Find the equation to the circle :
Whose radius is $\sqrt{a^2-b^2}$ and whose center is (-a, -b) .

Answer 1

Fact:

Equation of circle with centre $(h,k)$ and having radius $r$ is given by,

$(x-h{)}^2+(y-k{)}^2=r^2$

Here, centre is $(-a,-b)$ and radius is $\sqrt{a^2-b^2}$

hence equation of required circle is given by,

$[x-(-a){]}^2+[y-(-b){]}^2=(\sqrt{a^2-b^2}{)}^2$

$\Rightarrow (x+a{)}^2+(y+b{)}^2=a^2-b^2$

$\Rightarrow x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$

$\Rightarrow x^2+y^2+2ax+2by+2b^2=0$