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(A) The given line is $2x - y = 10$,which can be written as $y = 2x - 10$. The slope of this line is $m = 2$.The line perpendicular to this line will

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$10$

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(A) The given line is $2x - y = 10$,which can be written as $y = 2x - 10$. The slope of this line is $m = 2$.The line perpendicular to this line will have a slope $m' = -\frac{1}{2}$.For a parabola $y^2 = 4a(x - h)$,the tangent with slope $m$ touches at the point $(h + \frac{a}{m^2}, \frac{2a}{m})$.Here,$4a = 4 \implies a = 1$,$h = 9$,and $m = -\frac{1}{2}$.The point of contact $P$ is $(9 + \frac{1}{(-1/2)^2}, \frac{2(1)}{-1/2}) = (9 + 4, -4) = (13, -4)$.The circle is $x^2 + y^2 - 14x - 8y + 56 = 0$. The centre $C$ is $(-\frac{-14}{2}, -\frac{-8}{2}) = (7, 4)$.The distance $CP = \sqrt{(13 - 7)^2 + (-4 - 4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is ...........

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