Geometric Probability- Circle and two points

A point $P$ is chosen $0.5$ units away from the centre of a circle of diameter $2$. Now two points are chosen randomly on the circumference of the circle. What is the probability that the triangle formed by those two points and point $P$ encloses the centre of the circle?

• My insights are, this problem is equaivalento to finding two points on different side of a semicircle. Hence Probability is 0.5 – user106448 Nov 17 '13 at 7:16
• This is one of the many similar questions that answered this problem. – achille hui Nov 17 '13 at 7:53

The probability is $\frac14$.
First consider an apparently more general problem where we don't know the abscissa of $P$. We only know that $P$ has coordinate $(x,0)$ (with $0<x\leq\text{radius of the circle}$).
Let's call $A$ and $B$ the two random points. By symmetry we can assume that $A$ is in the upper half. The triangle $ABP$ contains the origin if and only if $B$ is in the arc starting from $(-1,0)$ and ending at $-A$ (going anticlockwise).
This is independent from $x$, so we could consider that $P$ is the point $(1,0)$ and since the uniform distribution is rotation invariant we can consider that the $P$ is also a random point on the circle.  