Peaches, peaches, peaches, peaches, peaches … I likely won’t water you

You and your assistant are planning to irrigate a vast circular garden, which has a radius of 1 furlong. However, your assistant is somewhat lackadaisical when it comes to gardening. Their plan is to pick two random points on the circumference of the garden and run a hose straight between them.

You’re concerned that different parts of your garden—especially your prized peach tree at the very center—will be too far from the hose to be properly irrigated.

On average, how far can you expect the center of the garden to be from the nearest part of the hose?

Without loss of generality, let's assume that we choose a coordinate system such that one end of the hose is located at the point $(1,0)$ and the other end is located at $(\cos \theta, \sin\theta)$ for some uniformly random $\theta \sim U(0,2\pi).$ The minimal distance from the origin, where the prized peach tree is, and the chord traced by the hose occurs along the ray that perpendicularly bisects the chord. We can arrive here either by calculus and finding $$\lambda^* = \arg\max \{ \| (1-\lambda + \lambda \cos \theta, \lambda \sin \theta ) \|_2 \mid 0\leq \lambda \leq 1 \} = \frac{1}{2}$$ or by spatial reasoning about Lagrange multipliers and optimizers occurring when contours and contraints are normal to one another or by any other means, I suppose. Whichever Feynman-esque path we take to arrive at the perpendicular bisector, we then see through some trigonometry that this minimal distance is this $$d(\theta) = \left| \cos \frac{\theta}{2}\right|,$$ as a function of the random $\theta.$

So, the average distance between the randomly placed hose and your precious peach tree is \begin{align*}\bar{d} &= \frac{1}{2\pi} \int_0^{2\pi} d(\theta) d\theta \\ &= \frac{1}{\pi} \int_0^\pi \cos \frac{\theta}{2} \,d\theta \\ &= \frac{2 \sin \frac{\pi}{2} - 2 \sin 0}{\pi} = \frac{2}{\pi}\approx 0.636519772\dots\end{align*} furlongs, which is about 420.3 feet. This doesn't seem too terribly bad, but given that the spread of an average peach tree is only about 20 feet (according to a quick Googling), your assistant's method is not expected to provide a large amount of water to your peaches.