Question

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

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

Answer 1

Answer: (B)

$2\sqrt{5}$

Equation of circle;

$16x^2+16y^2+48x-8y-43=0$

$\text{Center}=(-\frac{3}{2},\frac{1}{4})=\left(\text{h,k}\right)$

Equation of line,

$8x-4y+73=0$

$a=8,b=-4,c=73$

We know least distance between circle and line

$\theta =\left|\frac{ah+bk+c}{\sqrt{a^2+b^2}}\right|-\left|\frac{8\left(\frac{-3}{2}\right)-4\left(\frac{1}{4}\right)+73}{\sqrt{64+16}}\right|$

$\Rightarrow 3\sqrt{5}$

Subtract radius from this,

$3\sqrt{5}-r=3\sqrt{5}-\sqrt{5}=2\sqrt{5}$