Let the circle $x^{2} + y^{2} = 4$ intersect $x$-axis at the points $A(a,0),a 0$ and $B(b,0)$.

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Let the circle $x^{2} + y^{2} = 4$ intersect $x$-axis at the points $A(a,0),a > 0$ and $B(b,0)$. Let $P(2cos\alpha$, $2sin\alpha),0 < \alpha < \frac{\pi}{2}$ and $Q(2cos\beta,2sin\beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of AQ and BP lies on :

Accepted Answer

Let point of intersection $R(h,k)$$$\begin{matrix} & m_{BR} = m_{BP} \Rightarrow \frac{k}{h + 2} = \frac{2sin\alpha}{2cos\alpha + 2} \Rightarrow \frac{k}{h + 2} = tan\frac{\alpha}{2} \ & m_{AR} = m_{AQ} \Rightarrow \frac{k}{h - 2} = \frac{2sin\beta}{2cos\beta - 2} = \frac{sin\beta}{cos\beta - 1} = - cot\frac{\beta}{2} \ & \frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4} \ & tan\left( \frac{\alpha}{2} - \frac{\beta}{2} ight) = tan\frac{\pi}{4} = 1 \ & \frac{tan\frac{\alpha}{2} - tan\frac{\beta}{2}}{1 + tan\frac{\alpha}{2}tan\frac{\beta}{2}} = 1 \ & \frac{\frac{k}{h + 2} + \frac{h - 2}{k}}{1 + \left( \frac{k}{h + 2} ight)\left( \frac{2 - h}{k} ight)} = 1 \Rightarrow \frac{k^{2} + h^{2} - 4}{\frac{k(h + 2)}{4}} = 1 \ & \frac{h^{2} + k^{2} - 4}{4k} = 1 \ & x^{2} + y^{2} - 4y - 4 = 0 \end{matrix}$$

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