Placing a small circle within a large one, trying to maximize the circumference converage Assume you have a circle of radius $1$.
We want to place a smaller circle of radius $r<1$ inside, such that as much of the outer circle's circumference is contained inside the smaller one's area.

How should we place it (how far from the center of the outer circle should the center of the smaller circle be, as a function of $r$?).

What is the length of the covered circumference fraction?
 A: Place the small circle so that the two points where it intersects the big circle are diametrically opposite (with respect to the small circle). This is the biggest chord, and thus the biggest part of the circumference, that the small circle is able to cover.
A: Arc length = 2 x LargeRadius x Theta, where sin(Theta)= SmallRadius/LargeRadius
A: Let $\theta$ be the angle subtended by the circumference of the big circle (radius $=1$) intercepted by the smaller circle with radius $r(<1)$ then the length of intercepted circumference $$=\text{(radius)} \times \text{angle of aperture }=1\times \theta=\theta$$ 
Now, the length of common chord of small & big circles $$=2\sin\theta$$
But, the length of intercepted circumference ($\theta$) will be maximum if the common chord (of smaller & outer circles) is maximum i.e. the length of common chord is equal to the diameter $2r$ of smaller circle 
hence, equating the lengths of common chord, one should get 
$$2\sin \theta=2r$$$$\implies \sin\theta=r$$
Now, let $d$ be the distance between the centers of the circles & join the centers of circles to obtain a right triangle, 
Applying Pythagoras Theorem $$\cos\theta=\frac{d}{1}$$ 
$$d=\cos\theta$$ 
$$=\sqrt{1-\sin^2\theta}$$$$\color{}{=\sqrt{1-r^2}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{\text{distance between centers of circles}=\sqrt{1-r^2}}}$$
$\forall \ \ 0<r<1$  
