Just to reiterate Zilliput's observation: $\text{distance}\,(P, Q)$ denotes a function of two variables, one variable is the point $P = (x_1, y_1)$, which we can take to be the starting point, and the other is $Q = (x_2, y_2)$, which we can take to be the ending point. As a function, it takes as input the two points $P$ and $Q$, and returns (outputs) the real-valued $\text{distance}$ between inputs/points $P$ and $Q$.
Recall from your earlier question that $$\Delta x = x_2 - x_1,\;\;\Delta y = y_2 - y_1,$$ represent the change/displacement in coordinates when moving from $$P=(x_1, y_1) \longrightarrow Q = (x_2, y_2)$$
And so $$\text{distance}\,(P, Q) = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Now it can be seen more explicitly that $\;\text{distance}(P, Q)\;$ is a function of the two points, $P, Q$.
So in this context $(P, Q)$ is an ordered pair of inputs, or "arguments" which happen each to be points. That is not to say that $(P, Q)$ is an ordered pair representing the $x, y$-coordinates of a point in the plane.
In mathematics, an ordered pair can be an ordered pair of many entities, not just an ordered pair of coordinates in the Cartesian plane. We can define addition on say, natural numbers as a function $\;+(m, n) = m + n,\;$ where the ordered pair here is a pair of natural numbers, $m, n.\;$ Or, we can define the union of two sets as a function of an ordered pair of sets, $A, B:\;\;$ $\cup(A, B) = A \cup B$.
Unfortunately, the notation $(x, y)$ is, admittedly, ambiguous, because it could very well represent an ordered pair, but when we have $x \lt y$, might alternatively represent an interval:
E.g. $(1, 2): $ point in the plane?, or interval of the real line?
This is where context is crucial, and so its meaning can be discerned, usually, from context. But when we have a function, like distance, mapping an ordered pair of variables to some value, then we can "rule out" the "interval" interpretation.
Best Answer
The point at the centre of the diagram has coordinates $(x_1+x_2)/2$ and $(y_1+y_2)/2$. The distance $d$ from each of $P_1$ and $P_2$ to the centre is half the distance between $P_1$ and $P_2$, so $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}/2$. Then the distance $h$ from each of $T_1$ and $T_2$ to the centre follows by Pythagoras: $h=\sqrt{a^2-d^2}$. The vector with components $y_2-y_1,x_1-x_2$ is orthogonal to the line $P_1P_2$ and thus points along the line $T_1T_2$. Thus you just have to add $\pm h$ times a normalized version of that vector to the centre to get $T_1$ and $T_2$:
$$ \begin{eqnarray} T_{1,2} &=& \frac12\left({x_1+x_2\atop y_1+y_2}\right)\pm\frac{\sqrt{a^2-\frac14((x_2-x_1)^2+(y_2-y_1)^2)}}{\sqrt{(x_1-x_2)^2+(y_2-y_1)^2}}\left(y_2-y_1\atop x_1-x_2\right) \\ &=& \frac12\left({x_1+x_2\atop y_1+y_2}\right)\pm{\sqrt{\frac{a^2}{(x_2-x_1)^2+(y_2-y_1)^2}-\frac14}}\left(y_2-y_1\atop x_1-x_2\right) \end{eqnarray} $$