### Part 1: Distance Between the Centers of the Circles

1. Let the radius of the lower left circle be $r_1$ and the radius of the upper right circle be $r_2$. The centers of these circles are at $(r_1, r_1)$ and $(a - r_2, b - r_2)$ respectively.
2. The horizontal component of the distance between the centers is $x = a - r_1 - r_2$.
3. The vertical component of the distance between the centers is $y = b - r_1 - r_2$.
4. The total distance between the centers is given by the Pythagorean theorem:
$$\sqrt{x^2 + y^2} = \sqrt{(a - r_1 - r_2)^2 + (b - r_1 - r_2)^2}$$
5. Since the circles are tangent to each other, the distance between their centers is equal to the sum of their radii:
$$\sqrt{(a - r_1 - r_2)^2 + (b - r_1 - r_2)^2} = r_1 + r_2$$
6. Let $S = r_1 + r_2$. Then we have:
$$\sqrt{(a - S)^2 + (b - S)^2} = S$$
7. Squaring both sides, we get:
$$(a - S)^2 + (b - S)^2 = S^2$$
8. Expanding and simplifying:
$$a^2 - 2aS + S^2 + b^2 - 2bS + S^2 = S^2$$
$$a^2 + b^2 - 2(a + b)S + S^2 = 0$$
9. This simplifies to the quadratic equation:
$$S^2 - 2(a + b)S + (a^2 + b^2) = 0$$
10. Solving this quadratic equation using the quadratic formula $S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$S = \frac{2(a + b) \pm \sqrt{4(a + b)^2 - 4(a^2 + b^2)}}{2}$$
$$S = (a + b) \pm \sqrt{2ab}$$
11. Since $S$ must be less than both $a$ and $b$, we choose the negative root:
$$S = (a + b) - \sqrt{2ab}$$
$$S = (\sqrt{a} - \sqrt{b})^2$$

### Part 2: Locus of the Tangency Point

1. From the value of $S$, we have $x = a - S$ and $y = b - S$.
2. Let the distance from the tangency point to $AD$ be $p$ and the distance from the tangency point to $AB$ be $q$.
3. Using the corrected formulas:
$$p = \frac{r_1 x}{S} + r_1 = \frac{r_1 (a - S)}{S} + r_1$$
$$q = \frac{r_1 y}{S} + r_1 = \frac{r_1 (b - S)}{S} + r_1$$
4. It follows that:
$$q = p + (b - a)$$
5. The locus must be a line segment that is included in the equation above. To find the full locus, we must find the endpoints.
6. Note that $p$ and $q$ are both maximized when $r_1$ is maximized. Let $m = \min(a, b)$. Then $\max r_1 = \frac{m}{2}$ so:
$$\max p = \frac{m}{2} + \frac{r_1 (a - S)}{S} = \frac{m}{2} + \sqrt{2ab} - b$$
$$\max q = \frac{m}{2} + \frac{r_1 (b - S)}{S} = \frac{m}{2} + \sqrt{2ab} - a$$
7. Similarly, $p$ and $q$ are minimized when $r_2$ is maximized, which is when $r_1 = S - \frac{m}{2}$.

8. Since any point between the endpoints is in the locus, the locus consists of the line segment connecting:
$$\left( a - \frac{m}{2}, b - \frac{m}{2} \right) \text{ and } \left( \frac{m}{2} + \sqrt{2ab} - b, \frac{m}{2} + \sqrt{2ab} - a \right)$$

The final answer is $\boxed{ \left( \sqrt{a} - \sqrt{b} \right)^2 }$ for the distance between the centers and the line segment connecting $\left( a - \frac{m}{2}, b - \frac{m}{2} \right)$ and $\left( \frac{m}{2} + \sqrt{2ab} - b, \frac{m}{2} + \sqrt{2ab} - a \right)$ for the locus.