Let the circle x^{2}+y^{2}=4 intersect the x- | vedclass

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(A) The circle is $x^{2}+y^{2}=4$. The points of intersection with the $x$-axis are $A(2,0)$ and $B(-2,0)$.Let the point of intersection of $AQ$ and $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The circle is $x^{2}+y^{2}=4$. The points of intersection with the $x$-axis are $A(2,0)$ and $B(-2,0)$.Let the point of intersection of $AQ$ and $BP$ be $R(h,k)$.The slope of $BP$ is $m_{BP} = \frac{2 \sin \alpha - 0}{2 \cos \alpha - (-2)} = \frac{2 \sin \alpha}{2(\cos \alpha + 1)} = 	an \frac{\alpha}{2}$.The equation of line $BP$ is $y - 0 = 	an \frac{\alpha}{2} (x + 2)$.The slope of $AQ$ is $m_{AQ} = \frac{2 \sin \beta - 0}{2 \cos \beta - 2} = \frac{2 \sin \beta}{-2(1 - \cos \beta)} = -\cot \frac{\beta}{2}$.The equation of line $AQ$ is $y - 0 = -\cot \frac{\beta}{2} (x - 2)$.Given $\alpha - \beta = \frac{\pi}{2}$,so $\frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4}$.From the equations,$	an \frac{\alpha}{2} = \frac{k}{h+2}$ and $\cot \frac{\beta}{2} = -\frac{k}{h-2}$.Using $	an(\frac{\alpha}{2} - \frac{\beta}{2}) = 	an \frac{\pi}{4} = 1$,we get $\frac{	an \frac{\alpha}{2} - 	an \frac{\beta}{2}}{1 + 	an \frac{\alpha}{2} 	an \frac{\beta}{2}} = 1$.Substituting the values,$\frac{\frac{k}{h+2} + \frac{h-2}{k}}{1 + (\frac{k}{h+2})(\frac{2-h}{k})} = 1$.This simplifies to $\frac{k^2 + h^2 - 4}{4k} = 1$,which gives $h^2 + k^2 - 4k - 4 = 0$.Thus,the locus is $x^2 + y^2 - 4y - 4 = 0$.

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