Question

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

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

Answer 1

Answer: (B)

2$\sqrt{5}$

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

i.e., $x^2+y^2+3x-\frac{y}{2}-\frac{43}{16}=0$

Therefore center;$(\frac{-3}{2},\frac{1}{4})$

Radius=$\sqrt{\frac{9}{4}-\frac{1}{16}+\frac{43}{16}}=\sqrt{\frac{80}{16}}$

$=\sqrt{5}$

Perpendicular distance of center from line$=\frac{-12-1+73}{\sqrt{80}}$

$=\frac{60}{\sqrt{50}}=\frac{15}{\sqrt{5}}=3\sqrt{5}$

$\therefore$ Least distance of line from circle$=3\sqrt{5}-\sqrt{5}$

$=2\sqrt{5}$