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 $\_\_\_\_$ .
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Solution
$
y^2=4(x-9)
$
slope of tangent $=\frac{-1}{2}$
Point of contact P $\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)$
$
\mathrm{P}(13,-4)
$
center of circle $C(7,4)$
distance $\mathrm{CP}=\sqrt{(13-7)^2+(-4-4)^2}$
$
=10
$
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