Question

Tangent at $P(1,7)$ to the curve $y=x^2+6$ touches the circle $x^2+y^2+16x+12y+c=0$ at a point $Q$. Then the coordinates of $Q$ are

• A. $(-6,-11)$
• B. $(-9,-13)$
• C. $(1,-2)$
• D. $(-6,-7)$

Answer 1

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

Equation of tangents at $p(1,7)$ to $y=x^2+6$ is
slope $=y^{\prime }=2x$

The slope at $x=1$ is 2

Equation of tangent is ;
$\therefore (y-7)=2(x-1)$

$y=2x+5$ --(1)

$y=2x+5$ is also a tangent to $x^2+y^2+16x+12y+c=0$ at Q.

The length of the perpendicular from $P(x_1,y_1)$ on $ax+by+c=0$ is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\frac{(a{x}_1+b{y}_1+c)}{a^2+b^2}$

Foot of perpendicular from centre $(-8,-6)$ on $-2x+y-5=0$ is

$Q(h,k)$

$\frac{h+8}{-2}=\frac{k+6}{1}=-\frac{(16-6-5)}{5}$

$h=2-8$ and $k=-1-6$

$h=-6$ $k=-7$