Suppose we consider some hyperbolic circle with center $iz$ using the upper-half plane model of hyperbolic geometry, and in the interior we have a point $x+iy$. How does one prove that $iy$ is also part of the interior, as the picture seems to suggest? In other words, is $\text{dist}_{\mathbb{H}}(iy,iz) \leq \text{dist}_{\mathbb{H}}(x+iy,iz)$?
Yes. Note that the curve you drew between $iz$ and $x+iy$ does not well represent the geodesic arc between them. Let's see, first thing is that the geodesic center is at the geometric mean of the imaginary parts at maximum height and minimum height. This happens for a good reason: there are two ways to get arc-length parametrized geodesics:
$$ A + i e^t $$
$$ A + B \;\tanh t + i B \; \operatorname{sech} t, $$ where $A,B$ are real constants and $B > 0.$ The geodesic distance between two points is the absolute value of the difference in values of $t,$ once you have correctly picked out out the geodesic that passes through both. If two points have the same imaginary part, the geodesic has $A$ equal to the average of their real parts. If two points have the same real part, it is the first type, a vertical line. If the two points differ in both real and imaginary part, $A$ is where the perpendicular bisector of the segment between them hits that real axis; the curve is a semicircle when drawn, because $ \; \; \tanh^2 t + \; \mbox{sech}^2 \; t = 1. $
Well, that's a start. People do not seem to know this parametrization.
In case hyperbolic trig functions are a problem, $$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= $$ $$ \cosh^2 x - \sinh^2 x = 1 $$ $$ 1 - \tanh^2 x = \mbox{sech}^2 \; x $$ $$ \cosh x + \sinh x = e^x $$ $$ \cosh x - \sinh x = e^{-x} $$ $$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y $$ $$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y $$ $$ \tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y} $$ $$ \sinh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{2} } $$ $$ \cosh \frac{x}{2} = \sqrt { \frac{\cosh x + 1}{2} } $$ $$ \tanh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{\cosh x + 1} } = \frac{\cosh x - 1}{\sinh x} = \frac{\sinh x}{\cosh x + 1} $$ $$ \mbox{arcsinh} \; x = \log \left( x + \sqrt{x^2 + 1} \right) $$ $$ \mbox{arccosh} \; x = \log \left( x + \sqrt{x^2 - 1} \right), \; \; \; x \geq 1 $$ $$ \mbox{arctanh} \; x = \frac{1}{2} \; \log \left( \frac{1 + x}{1 - x} \right), \; \; \; |x | < 1 $$