The equation of the circle with radius 5 and | vedclass

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Question

(B) Since the circle touches both coordinate axes in the third quadrant,its center must be at a distance of $5$ units from both axes.In the third quad

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Since the circle touches both coordinate axes in the third quadrant,its center must be at a distance of $5$ units from both axes.In the third quadrant,both $x$ and $y$ coordinates are negative.Therefore,the center of the circle is $(-5, -5)$ and the radius is $r = 5$.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 with radius $5$ and touching the coordinate axes in the third quadrant is

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