Let O be the origin and OP,OQ be tangents to the circle x^2+y^2-6x+4y+8=0 at points P,Q on…

Question

Let $O$ be the origin and $OP,OQ$ be tangents to the circle $x^2+y^2-6x+4y+8=0$ at points $P,Q$ on it. If the circumcircle of triangle $OPQ$ passes through the point $\left(\alpha,\dfrac12\right)$, then a value of $\alpha$ is:

Accepted Answer

Circle centre $C=(3,-2)$. Since $\angle OPC=\angle OQC=90^\circ$, $P,Q$ lie on the circle with diameter $OC$:
$$x(x-3)+y(y+2)=0\ \Longrightarrow\ x^2+y^2-3x+2y=0.$$

Through $\left(\alpha,\dfrac12\right)$: $\alpha^2+\dfrac14-3\alpha+1=0\Rightarrow4\alpha^2-12\alpha+5=0\Rightarrow(2\alpha-1)(2\alpha-5)=0\Rightarrow\alpha=\dfrac12,\dfrac52.$

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