Find the standard form of the circle whose center is at (3,2) and which passes through the origin.

Standard form of the equation of a circle and graphs
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(x-3)^2+(y-2)^2=a^2 is the general form of the equation where a=radius.

If the circle passes through the origin (0,0) lies on the circle: 9+4=13=a^2, making the equation:

(x-3)^2+(y-2)^2=13 or (x-3)^2/13+(y-2)^2/13=1.

by Top Rated User (1.1m points)

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