Question 422641
x^2+y^2-4x+6y+9=0 find the radius

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Standard form for a circle:
(x-h)^2+(y-k)^2=r^2, (h,k) being the (x,y) coordinates of the center, and r=radius.

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x^2+y^2-4x+6y+9=0
completing the square,
(x^2-4x+4)+y^2+6y+9)=-9+4+9=4
(x-2)^2+(y+3)^2=4
This is a circle with center at (2,-3) with radius=sqrt(4)=2