Answer:
The radius is r=5 units
Step-by-step explanation:
we know that
The equation of the circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center and r is the radius
we have
[tex]x^{2}+y^{2}-12x+6y+20=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-12x)+(y^{2}+6y)=-20[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side
[tex](x^{2}-12x+36)+(y^{2}+6y+9)=-20+36+9[/tex]
[tex](x^{2}-12x+36)+(y^{2}+6y+9)=25[/tex]
Rewrite as perfect squares
[tex](x-6)^{2}+(y+3)^{2}=5^{2}[/tex]
therefore
The center is the point (6,-3) and the radius is r=5 units
