Let O be origin and OP and OQ be tangents to circle x^2+y^2-6 x+4 y+8=0 at point P and Q on it. If c

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10-1.Circle and System of Circles

hard

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6 x+4 y+8=0$ at the point $P$ and $Q$ on it. If the circumcircle of the triangle OPQ passes through the point $\left(\alpha, \frac{1}{2}\right)$, then a value of $\alpha$ is

A

$\frac{3}{2}$

B

$\frac{5}{2}$

C

$1$

D

$-\frac{1}{2}$

(JEE MAIN-2023)

Solution

Circumcircle of $\triangle OPQ$

$(x-0)(x-3)+(y-0)(y+2)=0$

$x^2+y^2-3 x+2 y=0$

$\text { passes through }\left(\alpha, \frac{1}{2}\right)$

$\therefore \alpha^2+\frac{1}{4}-3 \alpha+1=0$

$\Rightarrow \alpha^2-3 \alpha+\frac{5}{4}=0 \Rightarrow 4 \alpha^2-12 \alpha+5=0$

$\Rightarrow 4 \alpha^2-10 \alpha-2 \alpha+5=0$

$(2 \alpha-1)(2 \alpha-5)=0 \therefore \alpha=\frac{1}{2}, \frac{5}{2} \text { Ans. }$

Standard 11

Mathematics

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