Question

The tangent at $(1,7)$ to the curves $x^2=y-6x$ 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

The correct option is D (-6, -7)

The tangent at $(1,7)$ to the parabola $x^2=y-6x$ is
$x\left(1\right)=\frac{1}{2}(y+7)-6$
[replacing $x^2\to x{x}_1$ and $2y\to y+y_1$]
$\Rightarrow 2x=y+7-12$
$\Rightarrow y=2x+5$ ... (i)
which is also tangent to the circle
$x^2+y^2+16x+12y+c=0$
i.e. $x^2+(2x+5{)}^2+16x+12(2x+5)+c=0$ must have equal roots i.e., $\alpha =\beta$
$\Rightarrow 5x^2+60x+85+c=0$
$\Rightarrow \alpha +\beta =\frac{-60}{5}$
$\Rightarrow \alpha =-6$
$\therefore x=-6$ and $y=2x+5=-7$
$\therefore$ Point of contact is (-6, -7).