KCET · Maths · Circle
The area of the circle having its centre at \((3,4)\) and touching the line \(5 x+12 y-11=0\) is
- A \(16 \pi\) sq units
- B \(4 \pi\) sq units
- C \(12 \pi\) sq units
- D \(25 \pi\) sq units
Answer & Solution
Correct Answer
(A) \(16 \pi\) sq units
Step-by-step Solution
Detailed explanation
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
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
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