It may be convenient to get the constant "out of the way" by adding its opposite. Then the square can be completed for both x- and y-terms. The square is completed by adding the squares of half the coefficients of the linear terms.
[tex]x^{2}+y^{2}-10x-14y=26\\\\(x^{2}-10x+5^{2})+(y^{2}-14y+7^{2})=26+5^{2}+7^{2}\\\\(x-5)^{2}+(y-7)^{2}=100\\\\(x-5)^{2}+(y-7)^{2}=10^{2}[/tex]
We know the circle with center (h, k) and radius r will have the equation
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
By comparing the equation we have with the standard form equation for a circle, the center of the circle and its radius can now be read from the equation.
Center: (5, 7)
Radius: 10
