Circles Part 2

Concept:The circle $x^2+y^2=4$ has centre at origin and radius $2$. It meets the $x$-axis at $A(2,0)$ and $B(-2,0)$. We find the intersection of lines $AQ$ and $BP$, then use the given condition $\alpha-\beta=\pi/2$ to obtain the locus.Explanation:Let $P=(2\cos\alpha, 2\sin\alpha)$ and $Q=(2\cos\beta, 2\sin\beta)$. Let $R(h,k)$ be the intersection of $AQ$ and $BP$. Slope of $BP$ equals slope of $BR$: $\frac{k-0}{h-(-2)} = \frac{2\sin\alpha-0}{2\cos\alpha+2}$. Simplify: $\frac{k}{h+2} = \frac{\sin\alpha}{\cos\alpha+1} = \tan\frac{\alpha}{2}$. Thus $\boxed{\tan\frac{\alpha}{2} = \frac{k}{h+2}}$ ... (1). Slope of $QA$ equals slope of $QR$: $\frac{2\sin\beta-0}{2\cos\beta-2} = \frac{k-0}{h-2}$. Simplify: $\frac{\sin\beta}{\cos\beta-1} = \frac{k}{h-2}$. Using $\sin\beta = 2\sin\frac{\beta}{2}\cos\frac{\beta}{2}$ and $\cos\beta-1 = -2\sin^2\frac{\beta}{2}$, we get $-\cot\frac{\beta}{2} = \frac{k}{h-2}$. Hence $\tan\frac{\beta}{2} = \frac{2-h}{k}$ ... (2). Given $\alpha-\beta = \pi/2$, so $\frac{\alpha}{2} - \frac{\beta}{2} = \pi/4$. Take tangent: $\tan\left(\frac{\alpha}{2}-\frac{\beta}{2}\right) = 1$. Using formula: $\frac{\tan\frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 + \tan\frac{\alpha}{2}\tan\frac{\beta}{2}} = 1$. Substitute from (1) and (2): $\frac{\frac{k}{h+2} - \frac{2-h}{k}}{1 + \frac{k}{h+2}\cdot\frac{2-h}{k}} = 1$. Simplify numerator: $\frac{k}{h+2} + \frac{h-2}{k} = \frac{k^2 + (h-2)(h+2)}{k(h+2)} = \frac{k^2 + h^2 - 4}{k(h+2)}$. Denominator: $1 + \frac{2-h}{h+2} = \frac{(h+2)+(2-h)}{h+2} = \frac{4}{h+2}$. So the equation becomes $\frac{\frac{k^2+h^2-4}{k(h+2)}}{\frac{4}{h+2}} = 1$. This simplifies to $\frac{k^2+h^2-4}{4k} = 1$. Thus $k^2+h^2-4 = 4k$, i.e., $h^2 + k^2 -4k -4 = 0$. Replace $(h,k)$ with $(x,y)$: $x^2+y^2-4y-4=0$. Answer:Option D: $x^2+y^2-4y-4=0$.