Answer:
area of circle F = 9 pi
Explanation:
An interesting problem, combining geometry and algebra.
It has not been indicated, but it will be assumed that GH is a tangent to the circle F, hence the tangent-secant theorem from point H applies.
Produce HF to point E on the circumference such that EJ forms a diameter of circle F.
By the tangent-secant theorem,
HG^2 = HJ*HE
Substitute given values,
(x+2)^2 = 2*(2*(2x-1)+2)
Expand algebraically
x^2+4x+4 = 2(4x-2+2)
x^2+4x+4 = 8x
x^2-4x+4 = 0
Factor
(x-2)^2 = 0
=>
x=2
Substitute x=2 into radius FG = 2(x)-1 = 3
Therefore area of circle F
= pi r^2
= pi (3)^2
= 9 pi