Question 809582
{{{(x-x[0])^2+(y-y[0])^2=r^2}}} this is the standard equation for a circonference with center at point ({{{x[0]}}},{{{y[0]}}}) and radius r.
So let's try to put the given equation in standard form.
{{{x^2-6x+y^2+10y=2}}} 
{{{x^2-2*3*x+y^2-2*(-5)*y=2}}} 
{{{x^2-2*3*x+3^2-3^2+y^2-2*(-5)*y+(-5)^2-(-5)^2=2}}} 
{{{x^2-2*3*x+3^2+y^2-2*(-5)*y+(-5)^2=2+3^2+(-5)^2}}} 
{{{(x-3)^2+(y-(-5))^2=2+3^2+(-5)^2}}} 
{{{(x-3)^2+(y-(-5))^2=2+9+25}}} 
{{{(x-3)^2+(y-(-5))^2=36}}} 
{{{(x-3)^2+(y-(-5))^2=6^2}}} 
Therefore the given equation stands for a circonference with center in (3,-5) and radius 6