Let P be any point on the circle x^2+y^2-2x-1 | vedclass

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(A) Let the circle $S_1 \equiv x^2+y^2-2x-1=0$ have centre $C(1,0)$ and radius $r_1 = \sqrt{2}$.Let the circle $S_2 \equiv x^2+y^2-2x=0$ have centre $

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$2x^2+2y^2-4x+1=0$

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(A) Let the circle $S_1 \equiv x^2+y^2-2x-1=0$ have centre $C(1,0)$ and radius $r_1 = \sqrt{2}$.Let the circle $S_2 \equiv x^2+y^2-2x=0$ have centre $C(1,0)$ and radius $r_2 = 1$.Since $P$ lies on $S_1$,$PA$ and $PB$ are tangents to $S_2$ from $P$. Thus,$CA \perp PA$ and $CB \perp PB$.In $	riangle CAB$,$CA = CB = r_2 = 1$. The chord of contact $AB$ is perpendicular to $CP$.Let $M$ be the midpoint of $AB$. Since $	riangle CAB$ is an isosceles triangle with $CA=CB$,the circumcentre $O$ of $	riangle CAB$ lies on the line $CP$.Let $P = (1 + \sqrt{2}\cos	heta, \sqrt{2}\sin	heta)$. The line $CP$ has the equation $y = 	an	heta(x-1)$.The distance $CM$ from $C(1,0)$ to the chord $AB$ is given by $CM = \frac{r_2^2}{CP} = \frac{1^2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.The circumcentre $O(h,k)$ lies on $CP$ at a distance $R_{circum}$ from $C$. In $	riangle CAM$,$CA=1$ and $AM = \sqrt{1^2 - (1/\sqrt{2})^2} = 1/\sqrt{2}$.The circumradius $R_{circum} = \frac{CA^2}{2CM} = \frac{1^2}{2(1/\sqrt{2})} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Thus,the distance of $O(h,k)$ from $C(1,0)$ is $1/\sqrt{2}$.$(h-1)^2 + k^2 = (1/\sqrt{2})^2 = 1/2$.$h^2 - 2h + 1 + k^2 = 1/2 \Rightarrow h^2 + k^2 - 2h + 1/2 = 0$.Multiplying by $2$,we get $2h^2 + 2k^2 - 4h + 1 = 0$.Replacing $(h,k)$ with $(x,y)$,the locus is $2x^2 + 2y^2 - 4x + 1 = 0$.

Let $P$ be any point on the circle $x^2+y^2-2x-1=0$ and $C$ be its centre. Let $AB$ be the chord of contact of $P$ with respect to the circle $x^2+y^2-2x=0$. Then the locus of the circumcentre of the triangle $CAB$ is

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