Question

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

• A. $\frac{\sqrt{5}}{2}$ units
• B. $2\sqrt{5}$ units
• C. $3\sqrt{5}$ units
• D. $4\sqrt{5}$ units

Answer 1

The correct option is B $2\sqrt{5}$ units

For given circle, centre will be $(\frac{-3}{2},\frac{1}{4})$ and radius $r=\sqrt{5}$ units

Now, perpendicular distance from centre to the line $8x-4y+73=0$ is
$d=\left|\frac{8\left(\frac{-3}{2}\right)-4\left(\frac{1}{4}\right)+73}{\sqrt{8^2+(-4{)}^2}}\right|$
$\Rightarrow d=\left|\frac{60}{4\sqrt{5}}\right|=3\sqrt{5}>r$ units

So, line lies externally to the circle.
$\therefore$ least distance is $|d-r|=2\sqrt{5}$ units