The radius of the sphere $x^2+y^2+z^2=12 x+4 y+3 z$ is
The radius of the sphere $x^2+y^2+z^2=12 x+4 y+3 z$ is
$\frac{13}{2}$
13
26
52
Solution
Given equation of sphere is
$x^2+y^2+z^2-12 x-4 y-3 z=0$
$\therefore \quad$ Centre of sphere is $\left(6,2, \frac{3}{2}\right)$.
$\begin{gathered}
\therefore \text { Radius of sphere }=\sqrt{(6)^2+(2)^2+\left(\frac{3}{2}\right)^2} \\
=\sqrt{36+4+\frac{9}{4}}=\sqrt{\frac{169}{4}}=\frac{13}{2}
\end{gathered}$