In a circle of radius R two points are chosen at random(the points can be anywhere, either within the circle or on the boundary). For a fixed number c, lying between 0 and R, what is the probability that the distance between the two points will not exceed c?

kadejoset

kadejoset

Answered question

2022-07-23

What is this geometric probability
In a circle of radius R two points are chosen at random(the points can be anywhere, either within the circle or on the boundary). For a fixed number c, lying between 0 and R, what is the probability that the distance between the two points will not exceed c?

Answer & Explanation

phinny5608tt

phinny5608tt

Beginner2022-07-24Added 17 answers

Step 1
Basically you have to distinguish two cases: A circle with radius c around the first point lies completely inside the circle that is span by R. Then the probability is c 2 R 2 , but this only happens in ( R c ) 2 R 2 of the cases.
Step 2
Therefore the total probability will be c 2 ( R c ) 2 R 2 + K ( 1 ( R c ) 2 R 2 ) .
With K being the probability of the distance between the two points larger then c in case of the circle around the first point with radius c intersecting with the outter circle. And ( 1 ( R c ) 2 R 2 ) being the probability for this to happen.
In order to calculate K you need the formula for circle-circle intersections:
Step 2
Given two circles with radi R and r the formula for the intersectionarea A is:
A ( R , r , d ) = r 2 cos 1 ( d 2 + r 2 R 2 2 d r + R 2 ) cos 1 ( d 2 + R 2 r 2 2 d R ) 1 2 ( d + r + R ) ( d + r R ) ( d r + R ) ( d + r + R )
We will now integrate over this d.
K = d = R c d = R ( A ( R , r , d ) d 2 R 2 )
You have to multiply with d 2 R 2 , since the probability grows with d that the first point picked lies at a specific distance from the center.

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