Question 214271
Passes through the origin and is concentric with the circle x^2-6x+y^2-4y+4=0
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Find the center of this circle
x^2-6x+y^2-4y+4=0
x^2-6x+9 + y^2-4y+4 = 9
(x-3)^2 + (y-2)^2 = 3^2
That's a circle of radius 3 with a center at (3,2)
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The other circle has the same center (concentric).  Find its radius:
r^2 = 3^2 + 2^2
r^2 = 13
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--> (x-3)^2 + (y-2)^2 = 13
or, x^2-6x+y^2-4y=0