Question

Find the equation of the circle with the center $(-a,-b)$ and radius $√(a^2-b^2)$

Answer 1

The equation of a circle with a center $(h, k)$and radius $r$ is given as $(x - h)^2 + (y -k)^2 = r^2$

It is given that the center $(h, k) = (-a, -b)$ and radius $(r) =√(a^2-b^2).$

Therefore, the equation of the circle is

$(x+a)^2+ (y+b)^2= (√(a^2-b^2))^2$

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

$⇒ x^2++y^2+2ax+2by +2b = 0$ (Ans.)