Question 198771
2(Pi)x-2(Pi)r == (theta)x

Radius of larger circle is x, radius of smaller circle is r.

(r+3)=x

2(Pi)(r+3)-2(Pi)r == θ x

(2(Pi)(r+3))-2(Pi)r == (θ(r+3))

(2(Pi)(r+3))-2(Pi)r == (θ(r+3))

θ = (6(Pi))/(r+3)

(6(Pi))/(r+3) == (2(Pi)(r+3))-C

C == (2 (Pi) (6 + (6 r)+ r^2))/(3+r) == 2(Pi)r

solves to be:

r = -2

2(Pi)(-2+3) - 2(Pi)(-2) = 2(Pi)-((-)4(Pi)) = 2(Pi)+4(Pi) = 6(Pi)