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

Question

Let the circle $x^2 + y^2 = 4$ intersect x-axis at the points A$(a, 0)$, $a > 0$ and B$(b, 0)$. Let $P(2 \cos \alpha, 2 \sin \alpha)$, $0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$

Accepted Answer

The circle is ${x}^{2}+{y}^{2}={2}^{2}$ (radius $r=2$ ).

Intersection with $x$-axis: Setting $y=0$ gives ${x}^{2}=4\Rightarrow x=\pm 2$.

Since $a>0,A=(2,0)$.

Therefore, $B=(-2,0)$.

Points $P$ and $Q$ :

$P=(2\cos \alpha ,2\sin \alpha )$

$Q=(2\cos \beta ,2\sin \beta )$

Let point of intersection of $AQ$ .\& $BP$ be $R(h,k)$.

So, we need to find locus of points of intersection of $AQ\&BP$.

Slope of $BP=$ slope of $BR$

$\begin{gathered} \frac{k-0}{h-(-2)}=\frac{2\sin \alpha -0}{2\cos \alpha +2} \ \frac{k}{h+2}=\frac{\sin \alpha }{\cos \alpha +1}\Rightarrow \frac{k}{h+2}=\frac{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{2{\cos }^{2}\frac{\alpha }{2}} \ \frac{k}{h+2}=	an \frac{\alpha }{2}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i) \end{gathered}$

Similarly, slope of $QA=$ slope of $QR$

$\begin{gathered} \frac{2\sin \beta -0}{2\cos \beta -2}=\frac{k-0}{h-2} \ \frac{\sin \beta }{\cos \beta -1}=\frac{k}{h-2}\Rightarrow \frac{2\sin \frac{\beta }{2}\cos \frac{\beta }{2}}{-2{\sin }^{2}\frac{\beta }{2}}=\frac{k}{h-2} \ -\cot \frac{\beta }{2}=\frac{k}{h-2} \ \cot \frac{\beta }{2}=\frac{k}{2-h}\Rightarrow 	an \frac{\beta }{2}=\frac{2-h}{k} \end{gathered}$

Given,

$\begin{gathered} \alpha -\beta =\frac{\pi }{2}\Rightarrow \frac{\alpha }{2}-\frac{\beta }{2}=\frac{\pi }{4} \ 	an (\frac{\alpha }{2}-\frac{\beta }{2})=	an \frac{\pi }{4} \ \frac{	an \frac{\alpha }{2}-	an \frac{\beta }{2}}{1+	an \frac{\alpha }{2}	an \frac{\beta }{2}}=1 \end{gathered}$

Substitute

$	an \frac{\alpha }{2}=\frac{k}{h+2},\, 	an \frac{\beta }{2}=\frac{2-h}{k}$

$\begin{gathered} \frac{\frac{k}{h+2}-\frac{2-h}{k}}{1+(\frac{k}{h+2})(\frac{2-h}{k})}=1 \ \frac{k}{h+2}+\frac{h-2}{k}=1+(\frac{k}{h+2})(\frac{2-h}{k}) \ \frac{{k}^{2}+{h}^{2}-4}{(h+2)k}=\frac{4k}{(h+2)k} \ \frac{{k}^{2}+{h}^{2}-4}{(h+2)k}=\frac{4k}{(h+2)k} \ {k}^{2}+{h}^{2}-4=4k \ {k}^{2}+{h}^{2}-4k-4=0 \end{gathered}$

Replace $(h,k)	o (x,y)$

${x}^{2}+{y}^{2}-4y-4=0$

Solutions · 1