Q102 / 160 · TS EAMCET 2009 · Mathematics

The radius of the sphere $x^2+y^2+z^2=12 x+4 y+3 z$ is

  • Correct answer
    A
    $\frac{13}{2}$
  • B
    13
  • C
    26
  • D
    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 $\left(6,2, \frac{3}{2}\right)$. $\begin{gathered} \therefore \text { Radius of sphere }=\sqrt{(6)^2+(2)^2+\left(\frac{3}{2}\right)^2} \\ =\sqrt{36+4+\frac{9}{4}}=\sqrt{\frac{169}{4}}=\frac{13}{2} \end{gathered}$

Hence, the correct answer is option: 'A'.

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