Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point $P$. The…
Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is __________
Solution
$y^2=4(x-9)$
slope of tangent $=\frac{-1}{2}$
Point of contact $\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)$
$\begin{aligned} & P(13,-4) \\ & \text { center of circle } C(7,4) \\ & \text { distance } C P=\sqrt{(13-7)^2+(-4-4)^2} \\ & =10\end{aligned}$