Circle 5 Question 5
5. 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 the $\triangle P A B$ is
(a) $x^{2}+y^{2}+4 x-6 y+19=0$
(2009)
(b) $x^{2}+y^{2}-4 x-10 y+19=0$
(c) $x^{2}+y^{2}-2 x+6 y-29=0$
(d) $x^{2}+y^{2}-6 x-4 y+19=0$
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Answer:
Correct Answer: 5. (b)
Solution:
- For required circle, $P(1,8)$ and $O(3,2)$ will be the end point of its diameter.
$$ \begin{aligned} \therefore & (x-1)(x-3)+(y-8)(y-2) \\ \quad & =0 \\ \Rightarrow & x^{2}+y^{2}-4 x-10 y+19=0 \end{aligned} $$
