By the property of a circumscribed circle, the tangents drawn from an external point are equal. Thus:
\[AB + BC + CD + DE + EF + FA = 2 \times \text {Perimeter of Hexagon.}\]
Rearranging:
\[AB + CD + EF = BC + DE + FA.\]
Correct Answer: Proved
The equation of a circle which touches the straight lines $x + y = 2$, $x - y = 2$ and also touches the circle $x^2 + y^2 = 1$ is: