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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
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Verified Answer
The correct answer is:
$x^2+y^2-4 x-10 y+19=0$
$x^2+y^2-4 x-10 y+19=0$
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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