[tex]\text{Equation of the circle in standard form:}\\\\(x-h)^2+(y-k)^2=r^2\ \text{where}\\\\\text{(h; k) - coordinates of the center of the circle}\\\\\text{r-the radius}[/tex]
[tex]\text{We have:}\\\\x^2-6x+y^2+10y-2=0[/tex]
[tex]\text{Use:}\ (a\pm b)^2=a^2\pm2ab+b^2[/tex]
[tex]\underbrace{x^2-2\cdot x\cdot3+3^2}_{(x-3)^2}-3^2+\underbrace{y^2+2\cdot y\cdot5+5^2}_{(y+5)^2}-5^2-2=0\\\\(x-3)^2+(y+5)^2-9-25-2=0\\\\(x-3)^2+(y+5)^2-36=0\ \ \ \ |+36\\\\(x-3)^2+(y+5)^2=36[/tex]
[tex]\text{coordinates of the center of the circle (3; -5)}\\\\\text{a radius}\ r=\sqrt{36}=6[/tex]
