Let the circle C_1: x^2+y^2-2(x+y)+1=0 and C_ | vedclass

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(A) The equation of circle $C_1$ is $x^2+y^2-2x-2y+1=0$. The centre of $C_1$ is $(1,1)$.The equation of circle $C_2$ with centre $(-1,0)$ and radius $

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$2$

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(A) The equation of circle $C_1$ is $x^2+y^2-2x-2y+1=0$. The centre of $C_1$ is $(1,1)$.The equation of circle $C_2$ with centre $(-1,0)$ and radius $r=2$ is $(x+1)^2+(y-0)^2=2^2$,which simplifies to $x^2+2x+1+y^2=4$,or $x^2+y^2+2x-3=0$.The equation of the common chord is given by $S_1-S_2=0$:$(x^2+y^2-2x-2y+1) - (x^2+y^2+2x-3) = 0$$-4x-2y+4=0$$2x+y=2$.The line intersects the $y$-axis where $x=0$. Substituting $x=0$ into $2x+y=2$ gives $y=2$. Thus,point $P$ is $(0,2)$.The distance of $P(0,2)$ from the centre of $C_1(1,1)$ is $d = \sqrt{(1-0)^2+(1-2)^2} = \sqrt{1^2+(-1)^2} = \sqrt{2}$.The square of the distance is $d^2 = 2$.

Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $C_2$ be a circle having centre at $(-1,0)$ and radius $2$. If the line of the common chord of $C_1$ and $C_2$ intersects the $y$-axis at the point $P$,then the square of the distance of $P$ from the centre of $C_1$ is:

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