Question 708177
Complete the square for both variables and convert the equation to standard form; from this you can read the point (h,k) for center of the circle, and read the radius value.


Given:  x^2+ y^2-6x+4y-3=0
{{{(x^2-6x)+(y^2+4y)-3=0}}}
The missing square for the x part is (6/2)^2=3^2=9
The missing square for the y part is (4/2)^2=4


{{{(x^2-6x+9)+(y^2+4y+4)-3-9-4=0}}}
{{{(x-3)^2+(y+2)^2-16=0}}}
{{{(x-3)^2+(y+2)^2=16}}}
{{{highlight((x-3)^2+(y+2)^2=4^2)}}}


The center is at (3,-2) and radius is 4 units.