Question 103514
{{{x^2-10x+y^2+8y=-10}}}
The quickest way is to complete the square for both x and y. Then compare your equation to the equation of a circle and pick off the center of the circle and the radius. 
{{{(x^2-10x+A)=x^2+2B+B^2=(x+B)^2}}} Completing the square for x.
{{{(y^2+8y+C)=x^2+2D+D^2=(x+D)^2}}}Completing the square for y.
where {{{2B=-10}}} and {{{A=B^2}}} 
and {{{2D=8}}} and {{{C=D^2}}}
Let's solve for B and A.
{{{2B=-10}}}
{{{highlight(B=-5)}}}
{{{A=B^2}}}
{{{highlight(A=25)}}}
Let's solve for D and C.
{{{2D=8}}}
{{{highlight(D=4)}}}
{{{C=D^2}}}
{{{highlight(C=16)}}}
{{{x^2-10x+y^2+8y=-10}}}
{{{(x^2-10x+A)+(y^2+8y+C)=-10+A+C}}}Add A and C to both sides.
{{{(x+B)^2+(y+D)^2=-10+A+C}}}Substitute with the completed squares from above.
{{{(x-5)^2+(y+4)^2=-10+25+16}}}Substitute your calculated values for A,B,C,D.
{{{(x-5)^2+(y+4)^2=-10+25+16}}}
{{{(x-5)^2+(y+4)^2=31}}}
{{{(x-h)^2+(y-k)^2=R^2}}} The equation of a circle centered at (h,k) with a radius R.
In this case, 
(h,k)=(5,-4) and
{{{R^2=31}}}
{{{R=sqrt(31)}}}
{{{drawing( 300, 300, -2, 12, -12, 2,grid( 1 ),blue(circle( 5, -4, .2 )),green(circle( 5, -4, 5.567 )))}}}