Verified Answer :

naǫ

\((x-h)^2+(y-k)^2=r^2\)
(h,k) - the center
r - the radius

\(4x^2+4y^2+4x-16y-19=0 \ \ \ |\div 4 \\ x^2+y^2+x-4y-\frac{19}{4}=0 \\ x^2+x+\frac{1}{4}+y^2-4y+4-\frac{19}{4}-\frac{1}{4}-4=0 \\ (x+\frac{1}{2})^2+(y-2)^2-\frac{20}{4}-4=0 \\ (x+\frac{1}{2})^2+(y-2)^2-5-4=0 \\ (x+\frac{1}{2})^2+(y-2)^2-9=0 \\ (x+\frac{1}{2})^2+(y-2)^2=9 \\ (x+\frac{1}{2})^2+(y-2)^2=3^2\)

The center - \((-\frac{1}{2},2)\)
The radius - \(3\)

konrad509

\(4x^2+4y^2+4x-16y-19=0 \\ 4x^2+4x+1+4y^2-16y+16-36=0\\ (2x+1)^2+(2y-4)^2=36\)

center - \((-1,4)\\\)
\(r=\sqrt{36}=6\)

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