For the four circles M, N, O and P,the follow | vedclass

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(B) The centres of the circles are found by completing the square or using the formula $(-g, -f)$ for $x^2 + y^2 + 2gx + 2fy + c = 0$:Circle $M: x^2 +

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Square

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(B) The centres of the circles are found by completing the square or using the formula $(-g, -f)$ for $x^2 + y^2 + 2gx + 2fy + c = 0$:Circle $M: x^2 + y^2 = 1$,centre $M = (0, 0)$Circle $N: x^2 + y^2 - 2x = 0$,centre $N = (1, 0)$Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$,centre $O = (1, 1)$Circle $P: x^2 + y^2 - 2y = 0$,centre $P = (0, 1)$Now,calculate the lengths of the sides:$MN = \sqrt{(1-0)^2 + (0-0)^2} = 1$$NO = \sqrt{(1-1)^2 + (1-0)^2} = 1$$OP = \sqrt{(0-1)^2 + (1-1)^2} = 1$$PM = \sqrt{(0-0)^2 + (0-1)^2} = 1$Since all sides are equal to $1$ and the adjacent sides are perpendicular (slopes of $MN$ is $0$ and $NO$ is undefined),the figure is a square.

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