Let O be the origin and OP and OQ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $\left( {\alpha ,{1 \over 2}} \right)$, then a value of $\alpha$ is :
Solution
Centre $(3,-2)$
<br/><br/>Equation of circumcircle is
<br/><br/>$$
\begin{aligned}
& x(x-3)+y(y+2)=0 \\\\
& \Rightarrow x^2-3 x+y^2+2 y=0
\end{aligned}
$$
<br/><br/>Since $\left(\alpha, \frac{1}{2}\right)$ is on the circle
<br/><br/>$$
\begin{aligned}
& \text { So } \alpha^2-3 \alpha+\frac{1}{4}+1=0 \\\\
& \Rightarrow 4 \alpha^2-12 \alpha+5=0 \\\\
& \Rightarrow \alpha=\frac{12 \pm \sqrt{144-80}}{8} \\\\
& =\frac{12 \pm \sqrt{64}}{8}=\frac{12 \pm 8}{8} \\\\
& \alpha=\frac{20}{8}, \frac{4}{8}=\frac{5}{2}, \frac{1}{2}
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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