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 p…

Question

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 $$\left( {\alpha ,{1 \over 2}} \right)$$

Accepted Answer

Centre $(3,-2)$

Equation of circumcircle is

$$
\begin{aligned}
& x(x-3)+y(y+2)=0 \\
& \Rightarrow x^2-3 x+y^2+2 y=0
\end{aligned}
$$

Since $\left(\alpha, \frac{1}{2}\right)$ is on the circle

$$
\begin{aligned}
& \text { So } \alpha^2-3 \alpha+\frac{1}{4}+1=0 \\
& \Rightarrow 4 \alpha^2-12 \alpha+5=0 \\
& \Rightarrow \alpha=\frac{12 \pm \sqrt{144-80}}{8} \\
& =\frac{12 \pm \sqrt{64}}{8}=\frac{12 \pm 8}{8} \\
& \alpha=\frac{20}{8}, \frac{4}{8}=\frac{5}{2}, \frac{1}{2}
\end{aligned}
$$

Solutions · 1