Calculus 10th Edition

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