A circle is inscribed 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:

Options

A. 12
B. 11
C. 9
D. none of these

Answer

2

Explanation

`OA=1`
`r=OA cos 30^(@)=(sqrt(3))/(2)`
The equation of circle is
`x^(2)+y^(2)=(3)/(4)`
Let `P-= (x_(1),y_(2))`.
`PA^(2)+PB^(2)+PC^(2)+PD^(2)=x_(1)^(2)+(y_(1)-1)^(2)+(x_(1)+sqrt(3))^(2)+y_(1)^(2)+x_(1)^(2)+(y_(1)+1)^(2)+(x_(1)-sqrt(3))^(2)+y_(1)^(2)`
`=4x_(1)^(2)+4y_(1)^(2)+8=4(x_(1)^(2)+y_(1)^(2))+8`
`=4 xx (3)/(4)+8=11`