Let Gamma be a circle with diameter AB and ce | vedclass

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(B) Let the circle be $x^2 + y^2 = r^2$ with centre $O(0, 0)$.Let $M = (r \cos 	heta, r \sin 	heta)$.The tangent at $M$ is $x \cos 	heta + y \sin \

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a parabola

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(B) Let the circle be $x^2 + y^2 = r^2$ with centre $O(0, 0)$.Let $M = (r \cos 	heta, r \sin 	heta)$.The tangent at $M$ is $x \cos 	heta + y \sin 	heta = r$.The tangent $l$ at $B(r, 0)$ is $x = r$.Intersection $P$ of tangent at $M$ and $l$ is found by substituting $x = r$ into the tangent equation:$r \cos 	heta + y \sin 	heta = r \implies y \sin 	heta = r(1 - \cos 	heta) \implies y = r 	an(	heta/2)$.So,$P = (r, r 	an(	heta/2))$.The line through $P$ parallel to $AB$ (the $x$-axis) is $y = r 	an(	heta/2)$.The line $OM$ passes through $(0, 0)$ and $(r \cos 	heta, r \sin 	heta)$,so its equation is $y = x 	an 	heta$.$Q(h, k)$ is the intersection of $y = r 	an(	heta/2)$ and $y = x 	an 	heta$.Thus,$k = r 	an(	heta/2)$ and $k = h 	an 	heta$.Using $	an 	heta = \frac{2 	an(	heta/2)}{1 - 	an^2(	heta/2)}$,we have $k = h \cdot \frac{2(k/r)}{1 - (k/r)^2}$.$1 - k^2/r^2 = 2h/r \implies r^2 - k^2 = 2hr \implies k^2 = -2r(h - r/2)$.The locus of $Q(x, y)$ is $y^2 = -2r(x - r/2)$,which is a parabola.

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