$A(0,0),B(r,0),C(1,0),r=AB/AC$.
The semicircles with diameters $AB,AC,BC$ are tangent to a circle at $P_A,P_B,P_C$ respectively.
From MathWorld equation (4)-(9), the coordinates of $P_A,P_B,P_C$ are \begin{align} x_A &= \frac{r}{{(1-r)^2}} & y_A &= \frac{r(1-r)}{{(1-r)^2}} \tag1\\ x_B &= \frac{r(1+r)}{{1+r^2}} & y_B &= \frac{r(1-r)}{{1+r^2}} \tag2\\ x_C &= \frac{r^2}{{1-2r+2r^2}} & y_C &= \frac{r(1-r)}{{1-2r+2r^2}} \tag3 \end{align}
I think (1) is wrong: When $r=1$ denominator is zero but should have $P_A=C$.
I think the denominator in (1) should be changed to $1+(1-r)^2$.
Also, we should swap (2) and (3), since when $r=\frac12$ should have $x_B=y_B=\frac12$.
How to derive the correct formula?
You are correct. The coordinate of $P_A$ should read $$(x_A, y_A) = \left(\frac{r}{1 + (1-r)^2}, \frac{r(1-r)}{1 + (1-r)^2} \right).$$
The other coordinates for $P_B$ and $P_C$ are swapped as you have noted, although I suspect it is simply that the diagram is mislabeled and the intent was to have $P_C$ be the point of tangency with the the largest semicircle. I am a bit surprised to see this error; you should report it to MathWorld so they can correct it.
To derive these points, we can mechanically do it by solving for the intersection of two circles. For the first circle, choose $$(x - x_1)^2 + (y - y_1)^2 = r_1^2$$ where $(x_1, y_1)$ and $r_1$ are as described in Equations $(1)$-$(3)$ in the article for $n = 1$, and in sequence, $P_A$ is given by the intersection with the circle $$(x - r/2)^2 + y^2 = (r/2)^2;$$ $P_B$ is given by the intersection with the circle $$(x - 1/2)^2 + y^2 = (1/2)^2;$$ and $P_C$ is given by the intersection with the circle $$(x - (r+1)/2)^2 + y^2 = ((1-r)/2)^2,$$ if we follow the diagram in the article.
It may be possible to obtain the solution more generally and easily using inversive geometry.