A circle inscribed into a rhombus ABCD with one angle 60 o . 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 :

Step-by-step solution

In ∆ O D A tan 30 = 1 x x = 3 ∴  D = ( 3 , 0 ) r = 3 sin 3 0 ∘ = 3 2 PA 2 + PB 2 + PC 2 + PD 2 = x - 3 2 + y 2 + x 2 + y - 1 2 + x + 3 2 + y 2 + x 2 + y + 1 2 = 4 x 2 + y 2 + 2 = 4 x 2 + 2 ∵ x 2 + y 2 = r 2 = 4 3 4 + 2 = 1 1