Question

Tangent to the curve $y=x^2+6$ at a point $P($1, $7)$ 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. $(10,15)$
• D. $(-6,-7)$

Answer 1

Answer: (D)

$(-6,-7)$

The equation of parabola is $y=x^2+6$

Let the equation of tangent be $y=mx+c$

It touches the parabola , if $m^2-4(6-c)=0$

$\Rightarrow m^2+4c=24$

The point $P$ lies on a tangent , so we get $7=m+c$

By solving above two equations , we get $m=2$ and $c=5$

So the equation of tangent is $y=2x+5$

Let the coordinates of $Q$ be $(h,k)$

WE have $k=2h+5$

The center of circle is $(-8,-6)$ , so we have $\frac{k+6}{h+8}\times \left(2\right)=-1$

$\Rightarrow h+2k+20=0$

By solving above two equations , we get $h=-6,k=-7$