The equation of the circle of radius 5 and to | vedclass

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(D) Since the circle touches the coordinate axes in the $III$ quadrant,its center must be at $(-r, -r)$,where $r$ is the radius.Given the radius $r =

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$(x + 5)^2 + (y + 5)^2 = 25$

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(D) Since the circle touches the coordinate axes in the $III$ quadrant,its center must be at $(-r, -r)$,where $r$ is the radius.Given the radius $r = 5$,the center of the circle is $(-5, -5)$.The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = -5$,$k = -5$,and $r = 5$,we get:$(x - (-5))^2 + (y - (-5))^2 = 5^2$$(x + 5)^2 + (y + 5)^2 = 25$.

The equation of the circle of radius $5$ and touching the coordinate axes in the third quadrant is

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