Question

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

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

Answer 1

Answer: (D)

$(-6,-7)$

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