Answer
$(x+\frac{9}{2})^2+(y-1)^2+(z+5)^2=\frac{109}{4}$
Center: $\left(-\frac{9}{2},1,-5\right)$ and Radius:$\sqrt{\frac{109}{4}}=\frac{\sqrt{109}}{2}$
Work Step by Step
$x^2+y^2+z^2+9x-2y+10z+19=0$
$x^2+y^2+z^2+9x-2y+10z+19=(x+\frac{9}{2})^2-\frac{81}{4}+(y-1)^2-1+(z+5)^2-25+19=(x+\frac{9}{2})^2+(y-1)^2+(z+5)^2-\frac{109}{4}=0$
Therefore,
$(x+\frac{9}{2})^2+(y-1)^2+(z+5)^2=\frac{109}{4}$
Center: $\left(-\frac{9}{2},1,-5\right)$ and Radius:$\sqrt{\frac{109}{4}}=\frac{\sqrt{109}}{2}$
