The least distance of the line 8x-4y+73=0 from the circle 16x^2+16y^2+48x-8y-43=0 is

A) (square root of 5)/2 B) 2*square root of 5 C) 3*square root of 5 D) 4*square root of 5

Concept : The perpendicular distance is always minimum.
So if we calculate the perpendicular distance from centre of the circle to the line, and subtract the radius from it , we will get the distance of point on the circle to the line .

16x² + 16y² + 48x - 8y - 43 = 0
Divide both sides by 16 , we have
x² + y² + 3x - 1/2y = 43/16

So ( x + 3/2 )²  + ( y- 1/4)² = 43/16 + 9/4 + 1/16
Or ( x + 3/2 )²  + ( y- 1/4)² = 5
So the centre is (-3/2 , 1/4)  and radius = root(5) 

So the perpendicular distance from centre to the line is  :
$\frac{|A{x}_1 +B{y}_1 +C|}{\sqrt{A^2+B^2}} , ( here x_1 and y_1 are the point coordinates )$
And Ax + By +C = 0 is the line.

So the perpendicular distance = $\frac{|8\times {\displaystyle \frac{-3}{2}} -4\times {\displaystyle \frac{1}{4}} +73|}{\sqrt{8^2+(-4{)}^2}} =\frac{60}{\sqrt{80}}=\frac{15}{\sqrt{5}}$
So the minimum distance from any point on the line to the  circle is :
$\frac{15}{\sqrt{5}}-\sqrt{5}=\frac{10}{\sqrt{5}} =2\sqrt{5}$ (ans)