Question bank/JEE Main/Mathematics

A circle is inscribed (i.e. touches all four sides ) into a
rhombous ABCD with one angle 60°. 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:,d o`Ùk,d leprqHkqZt ABCD ftldk,d dks.k 60° vUrZ~fo"V gSA
o`Ùk Uæ dh lehiorhZ nwjh 1 gSA;fn o`Ùk ij dksbZ
fcUnq P gks] rks $|PA|^{2} + |PB|^{2} + |PC|^{2} + |PD|^{2}$dk eku
gksxk&

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$\because tan60^{\circ} = \frac{OA}{1} = \sqrt{3}$
$\therefore A(\sqrt{3},0)\ and\ C( - \sqrt{3},0)$
$\because sin60^{\circ} = \frac{r}{1} = \frac{\sqrt{3}}{2}$
Let coordinates of any point P on the circle be P
$\equiv (rcos\theta,rsin\theta)$
ekuk fd o`Ùk ij fLFkr fdlh fcUnq P
$(rcos\theta,rsin\theta)$
${PB}^{2} = (rcos\theta)^{2} + (1 - rsin\theta)^{2}$
${PC}^{2} = (rcos\theta + \sqrt{3})^{2} + (rsin\theta)^{2}$
$And\ {PD}^{2} = (rcos\theta)^{2} + (rsin\theta + 1)^{2}$
$\therefore{PA}^{2} + {PB}^{2} + {PC}^{2} + {PD}^{2} = 4r^{2} + 8 = 11\because r = \sqrt{3}/2$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx