The area of the circle having its centre at $(3,4)$ and touching the line $5 x+12 y-11=0$ is

• Correct answer
  A

  $16 \pi$ sq units
• B

  $4 \pi$ sq units
• C

  $12 \pi$ sq units
• D

  $25 \pi$ sq units

Solution

Given, centre of the circle is $(3,4)$. Here, $P C=$ Length of perpendicular to the line $5 x+12 y-11=0$ from the centre of the circle. $\begin{aligned} \Rightarrow \quad P C &=\frac{|5(3)+12(4)-11|}{\sqrt{25+144}}=\frac{|15+48-11|}{\sqrt{169}} \\ &=\left|\frac{52}{13}\right|=4 \end{aligned}$

$5 x+12 y-11=0$ $\therefore$ Radius of circle $=P C=4$ Hence, required area of circle $=\pi$ (radius) ${ }^{2}$ $=\pi(4)^{2}=16 \pi$ sq units

Hence, the correct answer is option: 'A'.