A circle is inscribed into a rhombus $ABCD$ with one angle $60^{\circ}$. The distance from the centr

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A circle is inscribed into a rhombus $ABCD$ with one angle $60^{\circ}$. The distance from the centre of the circle to the nearest vertex is equal to 1 . If $P$ is any point of the circle, then $|PA|^{2} + |PB|^{2} + |PC|^{2} + |PD|^{2}$ is equal to:

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[Key Concept: Incircle of a rhombus (angle $60^{\circ}$): the inradius and corner distances follow from the half-angle.]$r = \sqrt{3}sin30^{\circ} = \frac{\sqrt{3}}{2}$$$\begin{matrix} & \ (PA)^{2} + (PB)^{2} + (PC)^{2} + (PD)^{2} \ & \ = (x - \sqrt{3})^{2} + y^{2} + x^{2} + (y - 1)^{2} \end{matrix}$$$$\begin{matrix} & \ + (x + \sqrt{3})^{2} + y^{2} + x^{2} + (y + 1)^{2} \ & \ = 4\left( x^{2} + y^{2} + 2 ight) \ & \ = 4\left( \frac{3}{4} + 2 ight) = 11 \end{matrix}$$

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