Let RS be the diameter of the circle x^2+y^2= | vedclass

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(C) Given,$RS$ is the diameter of $x^2+y^2=1$. Let $P = (\cos 	heta, \sin 	heta)$.The tangent at $P$ is $x \cos 	heta + y \sin 	heta = 1$. The tan

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$A, C$

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(C) Given,$RS$ is the diameter of $x^2+y^2=1$. Let $P = (\cos 	heta, \sin 	heta)$.The tangent at $P$ is $x \cos 	heta + y \sin 	heta = 1$. The tangent at $S(1,0)$ is $x = 1$.Solving these,$Q = \left(1, \frac{1-\cos 	heta}{\sin 	heta}ight) = \left(1, 	an \frac{	heta}{2}ight)$.The line through $Q$ parallel to $RS$ (the $x$-axis) is $y = 	an \frac{	heta}{2}$.The normal at $P$ is $y = x 	an 	heta$.Let $E = (h, k)$. Then $k = 	an \frac{	heta}{2}$ and $k = h 	an 	heta$.Using $	an 	heta = \frac{2 	an(	heta/2)}{1 - 	an^2(	heta/2)}$,we get $k = h \frac{2k}{1-k^2}$.Thus,$1-k^2 = 2h$,or the locus is $2x = 1-y^2$.Checking the points:For $x = 1/3$,$1-y^2 = 2/3$ $\Rightarrow y^2 = 1/3$ $\Rightarrow y = \pm 1/\sqrt{3}$.Thus,points $(1/3, 1/\sqrt{3})$ and $(1/3, -1/\sqrt{3})$ lie on the locus.For $x = 1/4$,$1-y^2 = 2/4 = 1/2$ $\Rightarrow y^2 = 1/2$ $\Rightarrow y = \pm 1/\sqrt{2} 
eq \pm 1/2$.Therefore,the locus passes through points $A$ and $C$.

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