$\beta$ is placed on another circle with center point $\alpha$ such that it goes through an arbitrary point $\gamma$ on the circle with center point $\alpha$ . What is the probability that the circle with center point $\beta$ and the line segment $|\alpha \beta|$ intersect?

A circle with center point
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{4}$

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Call the other intersection point of the two circles $\gamma'$ . Now move $\gamma$ (and $\gamma'$ ) close enough to $\beta$ so that the circle with center $\beta$ goes through $\alpha$ . At that position, both circles have equal radii, and $\alpha \beta = \alpha \gamma = \beta \gamma$ so $\triangle \alpha \beta \gamma$ is equilateral making arc $\gamma \beta \gamma' = \frac{2\pi}{3}$ , i.e one third of the circumference of the circle with center $\alpha$ . Clearly for the circle with center $\beta$ to intersect $\alpha \beta, \gamma$ has to be on that third of the circumference, so the probability is $\boxed{\frac{1}{3}}$