MHT CET · Maths · Circle
The equation of the çircle which has its centre at the point \((3,4)\) and touches the line \(5 x+12 y-11=0\) is
- A \(x^2+y^2-6 x-8 y+9=0\)
- B \(x^2+y^2-6 x-8 y+25=0\)
- C \(x^2+y^2-6 x-8 y-9=0\)
- D \(x^2+y^2-6 x-8 y-25=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-6 x-8 y+9=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Radius }=\text { Distance of a point }(3,4) \text { form } \\
& 5 x+12 y-11=0 \\
& =\left|\frac{5(3)+12(4)-11}{\sqrt{25+144}}\right| \\
& =\left|\frac{15+48-11}{\sqrt{169}}\right| \\
& =\frac{52}{13} \\
& =4
\end{aligned}\)
\(\therefore \quad\) Required equation is
\(\begin{aligned}
& (x-3)^2+(y-4)^2=(4)^2 \\
& x^2+y^2-6 x-8 y+9=0
\end{aligned}\)
& \text { Radius }=\text { Distance of a point }(3,4) \text { form } \\
& 5 x+12 y-11=0 \\
& =\left|\frac{5(3)+12(4)-11}{\sqrt{25+144}}\right| \\
& =\left|\frac{15+48-11}{\sqrt{169}}\right| \\
& =\frac{52}{13} \\
& =4
\end{aligned}\)
\(\therefore \quad\) Required equation is
\(\begin{aligned}
& (x-3)^2+(y-4)^2=(4)^2 \\
& x^2+y^2-6 x-8 y+9=0
\end{aligned}\)
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