Point on the circle $x^2+y^2-2x+4y-4=0$  which is nearest to the line  $y=2x+11$ is :

Slope of line normal to circle and perpendicular to line
$m=-\frac{1}{2}=\tan \theta$
Co-ordinate of point lie on normal at a dist. Of 3 from center
$(1\pm 3\left(\frac{-2}{\sqrt{5}}\right),-2\pm 3\left(\frac{1}{\sqrt{5}}\right))$