For the four circles M, N, O and P, following four equations are...

$M: {x}^{2} + {y}^{2} = 1\text{center=}\left(0, 0\right)$

$N: {x}^{2} + {y}^{2} - 2x = 0, \text{center} = \left(1, 0\right)$

$O: {x}^{2} + {y}^{2} - 2x - 2y + 1 = 0, \text{center} = \left(1, 1\right)$

$P: {x}^{2} + {y}^{2} - 2y = 0, \text{center} = \left(0, 1\right)$

$MN = \sqrt{{\left(1 - 0\right)}^{2} + {\left(0 - 0\right)}^{2}} = 1$

$NO = \sqrt{{\left(1 - 1\right)}^{2} + {\left(0 - 1\right)}^{2}} = 1$

$OP = \sqrt{{\left(1 - 0\right)}^{2} + {\left(1 - 1\right)}^{2}} = 1$

$PM = \sqrt{{\left(0 - 1\right)}^{2} + {\left(1 - 1\right)}^{2}} = 1$

Then, all the sides are equal of a quadrilateral then $MNOP$ is a square.