The equation of the circle with center (-a, - | vedclass

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Question

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given center $(h, k) = (-a, -b)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given center $(h, k) = (-a, -b)$ and radius $r = \sqrt{a^{2}-b^{2}}$.Substituting these values into the formula:$(x - (-a))^{2} + (y - (-b))^{2} = (\sqrt{a^{2}-b^{2}})^{2}$$(x+a)^{2} + (y+b)^{2} = a^{2}-b^{2}$Expanding the squares:$x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} - b^{2}$Subtracting $a^{2}$ from both sides and adding $b^{2}$ to both sides:$x^{2} + y^{2} + 2ax + 2by + b^{2} + b^{2} = 0$$x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0$.

The equation of the circle with center $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$ is:

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