Question

A smallest distance the circle ${(x-5)}^2+{(y+3)}^2=1$ and the line $5x+12y-4=0$, is

• A. $\frac{1}{13}$
• B. $\frac{2}{13}$
• C. $\frac{3}{15}$
• D. $\frac{4}{15}$

Answer 1

The correct option is B $\frac{2}{13}$

$(x-5{)}^2+(y+3{)}^2=1\Rightarrow$ Equation of circle

Its centre $\Rightarrow (5,-3)$

Let $P$ be a point on line $5x+12y-4=0$

$\therefore PC=\left|\frac{25-36-4}{\sqrt{25+144}}\right|=\left|\frac{-15}{13}\right|=\frac{15}{13}$ (perpendicular distance)

$AP=PC-AC=\frac{15}{13}-1\frac{12}{13}$ [$\because AC=$ radius of circle $=1$]

$\therefore$ the smallest distance is $\frac{12}{13}unit$