Q34 / 90 · KVPY 2012 (SB/SX) · Mathematics
Suppose two perpendicular tangents can be drawn from the origin to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-2 \mathrm{py}+17=0$, for some real $\mathrm{p}$. then $|\mathrm{p}|=$
- A0
- B3
- Correct answerC5
- D17
Solution
$(x-3)^{2}+(y-p)^{2}=9-17+p^{2}$
Director circle is
$(x-3)^{2}+(y-p)^{2}=2\left(p^{2}-8\right)$
Passes through $(0,0)$
$\begin{array}{l}
9+p^{2}=2 p^{2}-16 \\
p^{2}=25 \Rightarrow p=\pm 5 \Rightarrow|p|=5
\end{array}$
Hence, the correct answer is option: 'C'.