If the common chord of the circles x^2+y^2-2x | vedclass

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(B) The equation of the common chord of the two circles $C_1: x^2+y^2-2x+2y+1=0$ and $C_2: x^2+y^2-2x-2y-2=0$ is given by $C_1 - C_2 = 0$. $(x^2+y^2-2

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$\left(1,-\frac{3}{4}\right)$

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(B) The equation of the common chord of the two circles $C_1: x^2+y^2-2x+2y+1=0$ and $C_2: x^2+y^2-2x-2y-2=0$ is given by $C_1 - C_2 = 0$. $(x^2+y^2-2x+2y+1) - (x^2+y^2-2x-2y-2) = 0$ $4y + 3 = 0 \Rightarrow y = -\frac{3}{4}$. Since this common chord is the diameter of circle $S$,the centre of circle $S$ must lie on the line $y = -\frac{3}{4}$. Among the given options,the points with $y$-coordinate $-\frac{3}{4}$ are $\left(\frac{1}{2}, -\frac{3}{4}\right)$,$\left(1, -\frac{3}{4}\right)$,and $\left(-\frac{1}{2}, -\frac{3}{4}\right)$. However,the centre of the circle $S$ must also satisfy the condition of being the midpoint of the chord if the chord is a diameter. The common chord is a horizontal line $y = -\frac{3}{4}$. The centre of the circle $S$ is $(1, -\frac{3}{4})$.

If the common chord of the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2-2x-2y-2=0$ is the diameter of a circle $S$,then the centre of the circle $S$ is

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