Without loss of generality, you can choose a coordinate system where the circle with radius $r$ is centered at $(1, 0)$.
If $r \ge 2$, it will cover the entire unit circle, and therefore have the total unit circle area, $\pi$. If $r \gt 1$, we can swap the situation around, simply by replacing $r$ with $1/r$ (scaling the situation by $1/r^2$ along each coordinate axis, so that the larger circle will have radius 1), and scale the result by $r^4$. So, we really only need to look at $0 \le r \le 1$.
We can describe the unit circle as $$y_0(x) = \pm\sqrt{1 - x^2}$$and the smaller circle as $$y_1(x) = \pm\sqrt{r^2 - (x - 1)^2}$$
These intersect at $y_0(x) = y_1(x)$, i.e. at
$$\left\lbrace\begin{aligned}
x &= 1 - \frac{r^2}{2} \\
y &= \pm\sqrt{r^2 - \frac{r^4}{4}} \\
\end{aligned}\right.$$
Note that the vertical line through the intersections is at $x \le 1$, so for $r \gt 0$, at most half of the $r$-radius circle intersects with the unit circle.
The vertical line through the intersections covers angle $\theta_0$ from the unit circle,
$$\theta_0 = 2\arctan\left(\frac{\sqrt{r^2 - \frac{r^4}{4}}}{1 - \frac{r^2}{2}}\right) = 2 \arctan\left(\frac{\sqrt{4 r^2 - r^4}}{2 - r^2}\right) \tag{1}\label{NA1}$$
and angle $\theta_1$ for the other circle (centered at $(1,0)$,
$$\theta_1 = 2\arctan\left(\frac{\sqrt{r^2 - \frac{r^4}{4}}}{\frac{r^2}{2}}\right) = 2\arctan\left(\sqrt{\frac{4}{r^2} - 1}\right)\tag{2}\label{NA2}$$
The area of the circular segment of the unit circle that is right of the intersections is $A_0$,
$$A_0 = \frac{1}{2}\left(\theta_0 - \sin\theta_0\right) \tag{3}\label{NA3}$$
and the area of the circular segment of the other circle left of the intersections is $A_1$,
$$A_1 = \frac{r^2}{2}\left(\theta_1 - \sin\theta_1\right) \tag{3}\label{NA4}$$
The total area $A$, as a function of $r$, is simply the sum of those two circular segments, connected at the vertical line between the two intersection points:
$$A = A_0 + A_1 \tag{5}\label{NA5}$$
You can simplify $\sin\theta_0$ to
$$\sin\theta_0 = \left(r - \frac{r^3}{2}\right)\sqrt{4 - r^2}$$
and $\sin\theta_1$ to
$$\sin\theta_1 = \frac{r}{2}\sqrt{4 - r^2}$$
which gives the entire formula as
$$\begin{aligned}
A(r) &= \arctan\left(\frac{\sqrt{4 r^2 - r^4}}{2 - r^2}\right) \\
\; &- \left(\frac{r}{2} - \frac{r^3}{4}\right)\sqrt{4 - r^2} \\
\; &+ \arctan\left(\sqrt{\frac{4}{r^2} - 1}\right) \\
\; &- \frac{r}{4}\sqrt{4 - r^2} \\
\end{aligned}$$
i.e.
$$A(r) = \arctan\left(\frac{\sqrt{4 r^2 - r^4}}{2 - r^2}\right) + \arctan\left(\sqrt{\frac{4}{r^2} - 1}\right) - \left(\frac{3 r - r^3}{4}\right)\sqrt{4 - r^2} \tag{6}\label{NA6}$$
