Question

A tangent at any point $P(1,7)$ on the parabola $y=x^2+6$, which is touching to the circle $x^2+y^2+16x+12y+c=0$ at point $Q$, then $Q$ is-

• A. $(-6,-7)$
• B. $(-10,-15)$
• C. $(-9,-7)$
• D. $(-6,-3)$

Answer 1

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

$y=x^2+6$

Tangent drawn at pt. $P(1,7)$

$\Rightarrow y=2x=2$ = slope of tangent at $P(1,7)$

eqn. of tangent : $y-7=2(x-1)$

$\Rightarrow y=2x+5$

or $2x-y+5=0$ __ (1)

eqn. of circle : $x^2+y^2+16x+12y+c=0$

$\Rightarrow (x+8{)}^2+(y+6{)}^2+c-64-36=0$

$\Rightarrow (x+8{)}^2+(y+6{)}^2+c-100=0$

$\Rightarrow$ circle is centred at (-8,-6)

$\Rightarrow (-g,-f)=(-8,-6)$

$\Rightarrow g=8,f=6$

Eqn. of circle is of the form of :

$x^2+y^2+2gx+2fy+c=0$

where $g=8,f=6$

Then eqn. of tangent to this circle at $Q(x_1,y_1)$

is $x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$

or $x(x_1+g)+y(y_1+f)+g{x}_1+f{y}_1+c=0$

Comparing it to eqn of tangent of parabola.

$2x-y+5=0$

$\Rightarrow x_1+g=2$

$y_1+f=-1$

$\Rightarrow x_1=2-g=-6$

$y_1=-1-f=-7$

$\Rightarrow Q(x_1,y_1)=(-6,-7)$