Question

Tangent to the parabola $y=x^2+6$ at $(1,7)$ touches the circle $x^2+y^2+16x+12y+c=0$ at the point

• A. $(-6,-9)$
• B. $(-13,-9)$
• C. $(-6,-7)$
• D. $(13,7)$

Answer 1

The correct option is C $(-6,-7)$

Equation of tangent at $(1,7)$ to $y=x^2+6$ is
$\frac{1}{2}(y+7)=x.1+6\Rightarrow y=2x+\mathrm{5...}\left(i\right)$
This tangent also touches the circle
$x^2+y^2+16x+12y+c=\mathrm{0....}\left(ii\right)$
Now solving Eqs (i) and (ii) we get
$\Rightarrow x^2+{(2x+5)}^2+16x+12(2x+5)+c=0$
$\Rightarrow 5x^2+60x+85+c=\mathrm{0....}\left(iii\right)$
Since roots are equal
$\Rightarrow b^2-4ac=0\Rightarrow {\left(60\right)}^2-4\times 5\times (85+c)=0$
$\Rightarrow 85+c=180$
On putting this value in Eq $\left(iii\right)$, we get
$5x^2+60x+180=0$
$\Rightarrow x=\frac{-60}{10}=-6$
Then from Eq $\left(i\right)$ $y=-7$

Hence, the point of intersection is $(-6,-7)$.