# Probability that a random chord intersects a circle.

Assume we got a big circle with radius $R$. Another circle with radius $r$ is uniformly distributed inside the big ones. Now I want to calculate the probability that a randomly distributed chord of the big circle intersects the small circle.

I have an idea: Assigned $D$ as the distance between the center of the small circle to the chord and $D$ is also followed an uniform distribution of $(0,2R)$. However I am not sure about it.

Edit : Sorry for the inconvenience. The small circle need to be inside the larger one. And for the chord, it was created by selecting two point randomly on the perimeter of the big circle.

• In order to answer this you need to be a little more specific about the distributions of your objects. For the smaller circle, must it be completely inside the larger circle, or can it be only partially inside? For the chord, what do you mean specifically when you say it is "randomly distributed"?
– Ben
May 20, 2018 at 6:07
• Sorry for the inconvenience. The small circle need to be inside the larger one. And for the chord, it was created by selecting two point randomly on the perimeter of the big circle. May 20, 2018 at 6:39
• 'Random chords' are tricky. Check out Bertrand's paradox. May 20, 2018 at 7:07
• @RogerJBarlow I think OP has chosen the $$random endpoints$"$ method. May 20, 2018 at 7:58
• @Alex Vong Yes I chose the random endpoints. I already looked at the Bertrand problem. I though I could figure out the probability by estimate the angle between two perpendicular lines from one endpoint to the small circle. However since the location of the circle is random, it's highly impossible to estimate it. May 20, 2018 at 8:28

The treatment of this problem is heavily marred by boundary effects. I suggest you replace the small disc by a point which is uniformly distributed in a large disc of radius $R-r$. As a countermeasure replace the chords with endpoints $R(\cos\alpha,\pm\sin\alpha)$ $\bigl(0\leq\alpha\leq{\pi\over2}\bigr)$ by parallel strips of width $2r$. You then have to compute the area of the intersection of such a strip with the disc of radius $R-r\,$; then finally integrate over $\alpha$.