Mathematics - Circle Question with Solution | TestHub

$AP=AQ=AM$ So $\angle PMQ=90°$
Locus of $M$ is a circle having $PQ$ as its diameter
Hence, $E_1: x^2+y^2-2y-39=0, x\ne \pm 2$
Clearly $\left(-2,7\right)$ is not included in the locus because we can not consider point circles, and $\left(0,\frac{3}{2}\right)$ does not lie on $E_1$ as it does not satisfy the equation above.

The equation of chord with $B\left(h,k\right)$ as mid point will be given by $T=S_1$ or

$xh+yk-y-k-39=h^2+k^2-2k-39$

or $xh+yk-y+k=h^2+k^2$

It passes through $R\left(1,1\right)$. So,

$h+k-1+k=h^2+k^2$

or the locus of $E_2$ is $x^2+y^2-x-2y+1=0$

But if $P$ is not the part of locus of $M$ then $PQ$ is not the chord of locus of $M$. So point $\left(\frac{4}{5},\frac{7}{5}\right)$ does not lie in $E_2$. $\left(\frac{1}{2},1\right)$ does not satisfy the equation hence does not lie in $E_2$.