The area of equilateral triangle inscribed in the circle \[x^2+y^2-4x+2y+1=0\] is of the form \[\frac{a√b}{c}\]
Find \[a+b+c\]
This is a circle centered at (2, -1) with a radius of 2
The side^2, S^2, of the triangle can be found as
S^2 = 2^2 + 2^2 - 2 ( 2 * 2) cos (120 degrees)
S^2 = 8 - 8 (-1/2)
S^2 = 8 + 4 = 12
The area of this triangle =
(1/2) S^2 sin (60 degrees) =
(1/2) 12 * sqrt (3) / 2 =
6 sqrt (3) / 2 = 3 sqrt (3) / 1
a + b + c = 3 + 3 + 1 = 7