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 __________

Find the distance of the point of tangency $$P$$ (where a line perpendicular to $$2x - y = 10$$ touches the parabola $$y^2 = 4(x - 9)$$) from the centre of the circle $$x^2 + y^2 - 14x - 8y + 56 = 0$$.

The line $$2x - y = 10$$ has slope $$2$$; hence a line perpendicular to it has slope $$m = -\tf\frac{1}{2}$$.

The parabola $$y^2 = 4(x - 9)$$ has vertex $$(9,0)$$ and $$4a = 4$$, so $$a = 1$$. Differentiating implicitly gives $$2y\,\f\frac{dy}{dx} = 4$$ and thus $$\f\frac{dy}{dx} = \f\frac{2}{y}$$. Setting this equal to $$-\tf\frac{1}{2}$$ yields $$\f\frac{2}{y} = -\tf\frac{1}{2}$$, so $$y = -4$$. Substituting $$y = -4$$ into the parabola equation gives $$(-4)^2 = 4(x - 9)$$, i.e.\ $$16 = 4(x - 9)$$ and hence $$x = 13$$. Therefore the point of tangency is $$P(13,-4)$$.

Rewriting the circle equation $$x^2 + y^2 - 14x - 8y + 56 = 0$$ by completing the square yields $$(x - 7)^2 + (y - 4)^2 = 9$$, so its centre is $$(7,4)$$.

The distance from $$P(13,-4)$$ to $$(7,4)$$ is $$$ d = \sqrt{(13 - 7)^2 + (-4 - 4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10. $$$

The distance is 10.