# Three points $P,Q,R$ on a circle $O$. Draw circles with diameters $PQ,QR, RP$. Probability when $O$ lies inside the three circles drawn?

Sample three random points $$P,Q,R$$ on a unit circle $$O$$. Draw circles with diameters $$PQ,QR,RP$$. Find the probability when origin $$O$$ lies inside at least one of the three circles drawn?

There is a well know problem where we are simply looking for probability when $$O$$ is within $$\triangle PQR$$ and the probability is $$\frac{1}{4}$$. but is there a solution when we are looking for the probability when origin lies within circles with diameters $$PQ,QR,RP$$?

Probability that the center of a circle lies within three chosen points on a circle

What is the probability that the center of the circle is contained within the triangle?

## 1 Answer

We will first look at probability of $$O$$ not falling in any of the $$3$$ circles.

As we are taking all $$3$$ segments as diameter of $$3$$ circles, their radii (say, $$r$$ for any one of them) has to be less than the perp distance from $$O$$ to the segment (say, $$h$$) for $$O$$ not to be in any of the circles.

Now, $$r^2 + h^2 = 1^2$$ given unit circle. As the radius has to be less than the distance from $$O$$ to the segment, $$r \leq h \implies r \leq \frac{1}{\sqrt 2} (\leq 45^0$$ angle).

So the segment should subtend an angle of $$\leq \frac{\pi}{2}$$ at $$O$$.

If we take point $$P$$ randomly on the unit circle, the probability of the next point $$Q$$ being within angle $$\frac{\pi}{2}$$ to it (on either side of it) will be $$\frac{\pi}{2\pi} = \frac{1}{2}$$.

Say points $$P$$ and $$Q$$ subtend an angle of $$\theta (\leq \frac{\pi}{2})$$ at $$O$$. Then point $$R$$ can be within $$(\frac{\pi}{2} - \theta)$$ angle on either side of $$P$$ or $$Q$$. But it can also fall between point $$P$$ and $$Q$$ hence the probability of point $$R$$ satisfying the condition such that point $$O$$ is not within the $$3$$ circles $$= \displaystyle \frac{2(\frac{\pi}{2} - \theta) + \theta}{2\pi} = \frac{\pi - \theta}{2\pi}$$ where $$0 \lt \theta \leq \frac{\pi}{2}$$.

The probability for point $$R = \frac{1}{\pi/2} \int_0^{\pi/2} (\frac{1}{2} - \frac{\theta}{2\pi}) d\theta = \frac{2}{\pi} (\frac{\pi}{4} - \frac{\pi}{16}) = \frac{3}{8}$$.

So probability that none of the circles have point $$O = \frac{1}{2} \times \frac{3}{8} = \frac{3}{16}$$.

So desired probability $$= 1 - \frac{3}{16} = \frac{13}{16}$$