Let the circle x^2+y^2=4 intersect the x-axis at the points A(a,0), a0, and B(b,0).

Question

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

Accepted Answer

$A=(2,0),\ B=(-2,0)$. Let $R(h,k)$ be the intersection of $AQ$ and $BP$. From the collinearity of $B,P,R$:

$$\frac{k}{h+2}=\tan\frac{\alpha}{2}, \qquad \text{and from } A,Q,R:\ \tan\frac{\beta}{2}=\frac{2-h}{k}.$$

Using $\dfrac{\alpha}{2}-\dfrac{\beta}{2}=\dfrac{\pi}{4}$ and the tangent subtraction formula:

$$\frac{\tan\frac{\alpha}{2}-\tan\frac{\beta}{2}}{1+\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}=1 \;\Rightarrow\; \frac{k}{h+2}+\frac{h-2}{k}=1+\frac{k}{h+2}\cdot\frac{2-h}{k}$$

which simplifies to $k^2+h^2-4=4k$, i.e. $x^2+y^2-4y-4=0$.

Solutions · 1