Q62 / 120 · VITEEE 2012 · Mathematics

The tangent at $(1,7)$ to the curve $x^{2}=y-6$ touches the circle $x^{2}+y^{2}+16 x+12 y+c=0$ at

  • A
    $(6,7)$
  • B
    $(-6,7)$
  • C
    $(6,-7)$
  • Correct answer
    D
    $(-6,-7)$

Solution

The tangent at $(1,7)$ to the curve $\mathrm{x}^{2}=\mathrm{y}-6$ is $$ \begin{array}{l} x=\frac{1}{2}(y+7)-6 \\ \Rightarrow 2 x=y+7-12 \\ \Rightarrow y=2 x+5 \end{array} $$ which is also tangent to the circle $x^{2}+y^{2}+16 x+12 y+c=0$ i.e. $x^{2}+(2 x+5)^{2}+16 x+12(2 x+5)+c=0$ $\Rightarrow 5 x^{2}+60 x+85+c=0$, which must have equal roots. Let $\alpha$ and $\beta$ are the roots of the equation. Then $\alpha+\beta=-12 \Rightarrow \alpha=-6$ $(\because \alpha=\beta)$ $\therefore x=-6, y=2 x+5=-7$ $\therefore$ Point of contact is $(-6,-7)$.

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

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