we know that
the standard form of the equation of the circle is
\((x-h)^{2} +(y-k)^{2}=r^{2}\)
where
(h,k) is the center of the circle
r is the radius of the circle
In this problem we have
\(x^{2} +y^{2} +8x-6y+21=0\)
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\((x^{2}+8x) +(y^{2}-6y)=-21\)
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
\((x^{2}+8x+16) +(y^{2}-6y+9)=-21+16+9\)
\((x^{2}+8x+16) +(y^{2}-6y+9)=4\)
Rewrite as perfect squares
\((x+4)^{2} +(y-3)^{2}=4\)
\((x+4)^{2} +(y-3)^{2}=2^{2}\)
the center of the circle is \((-4,3)\)
the radius of the circle is \(2\ units\)
therefore
the answer is
the radius of the circle is \(2\ units\)
