Let $A\_r = \{z\in\mathbb{C}: |z|\leq 1\}\cup\{z\in\mathbb{C}: |z-1|\leq r\}$ be the unit disk with a small "bump" (I'm interested in the regime $r\to 0$). What can be said about the logarithmic capacity of $A\_r$? I can prove that there exists a constant $\gamma>0$ such that $\mathop{\mathrm{cap}}(A\_r)=\gamma r^2 + $terms of smaller order, but have no idea about how to obtain the value of $\gamma$.


Related to the above question: is there any useful explicit form of the conformal mapping of $\{z\in\mathbb{C}: \Im(z)\leq 0\}$ to $\{z\in\mathbb{C}: \Im(z)\leq 0\}\cup \{z\in\mathbb{C}: |z|\leq 1\}$?



Capacity of a unit disk with a small bump
Both questions have exact explicit answer, which is explained in any good textbook on analytic functions. A region on the Riemann sphere is called a digon if its boundary consists of
two arcs of circles. Let $a$ and $b$ be the common endpoints of these arcs. 
Then
$$f(z)=\frac{z-a}{z-b}$$
sends $a,b$ to $0,\infty$, and the arcs to rays. So the image of our region is a sector. The interior angle at a corner of the digon be $\pi\alpha$. Then the interior angle at the vertex of the sector is also $\pi\alpha$. Next $f\_1(z)=z^{1/\alpha}$ maps this sector onto a half-plane. And the final ingredient
is conformal map of this half-plane onto the unit disk.


EDIT 2. Asymptotics is
$\mathrm{cap}(A\_r)=1+r^2/2+O(r^3),$ as $r\to 0$.
The exact formula is a bit complicated:
$$\frac{1}{\mathrm{cap}(A\_r)}=\frac{1}{4b}(1+\phi)(1+b^2)\sin\left(\frac{4}{1+\phi}\arctan b\right),$$
where
$$\phi=\frac{2}{\pi}\arcsin(r/2),$$
and
$$b=\frac{r}{2\sqrt{1-r^2/4}}.$$


