JEE Advanced · Mathematics · 12. Circle
Tangents drawn from the point \(P(1,8)\) to the circle \(x^2+y^2-6 x-4 y-11=0\) touch the circle at the points \(A\) and \(B\). The equation of the circumcircle of \(\triangle P A B\) is
- A
\(x^2+y^2+4 x-6 y+19=0\)
- B
\(x^2+y^2-4 x-10 y+19=0\)
- C
\(x^2+y^2-2 x+6 y-29=0\)
- D
\(x^2+y^2-6 x-4 y+19=0\)
Answer & Solution
Correct Answer
(B)
\(x^2+y^2-4 x-10 y+19=0\)
Step-by-step Solution
Detailed explanation
For required circle, \(P(1,8)\) and \(O(3,2)\) will be the end point of its diameter.
\[
\begin{aligned}
& \therefore(x-1)(x-3)+(y-8)(y-2)=0 \\
& \Rightarrow \quad x^2+y^2-4 x-10 y+19=0
\end{aligned}
\]
\[
\begin{aligned}
& \therefore(x-1)(x-3)+(y-8)(y-2)=0 \\
& \Rightarrow \quad x^2+y^2-4 x-10 y+19=0
\end{aligned}
\]
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