Q69 / 120 · VITEEE 2009 · Mathematics

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

  • Correct answer
    A
    $13 / 2$
  • B
    13
  • C
    26
  • D
    52

Solution

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

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

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