Question

find equation of circle whose radius is under root a^2-b^2 whose centre is (-a,-b)

Answer 1

$x^{2} + y^{2}+ 2ax +2by +2b^{2} = 0$

stepn by step

we can say that the locus of the point which is at $\sqrt{a^{2}- b^{2} }$

distance from the point (-a,-b) is the circle.

Hence by the distance formula we get

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

by solving we get

$x^{2} + y^{2}+ 2ax +2by +2b^{2} = 0$