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)

Answer 2

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