we know that
the standard form of the equation of the circle is
[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center of the circle
r is the radius of the circle
In this problem we have
[tex]x^{2} +y^{2} +8x-6y+21=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}+8x) +(y^{2}-6y)=-21[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+8x+16) +(y^{2}-6y+9)=-21+16+9[/tex]
[tex](x^{2}+8x+16) +(y^{2}-6y+9)=4[/tex]
Rewrite as perfect squares
[tex](x+4)^{2} +(y-3)^{2}=4[/tex]
[tex](x+4)^{2} +(y-3)^{2}=2^{2}[/tex]
the center of the circle is [tex](-4,3)[/tex]
the radius of the circle is [tex]2\ units[/tex]
therefore
the answer is
the radius of the circle is [tex]2\ units[/tex]
