Find the midpoint between two points on the circle

WLOG, we can let the circle be centered at O(0, 0) with radius = r.

Therefore, the equation of the circle is $x^2 + y^2 = r^2$

M(p, q) is point on this circle implies $p^2 + q^2 = r^2$ ……… (1)

By midpoint formula, $N(r, s) = N(\dfrac {x_1 + x_2}{2}, \dfrac {y_1+ y_2}{2})$

N(r, s) is a point on OK, the line perpendicular to $P_1P_2$. By two-point form, the equation of OK is

$y = \dfrac {y_1 + y_2}{x_1 + x_2}x$

M(p, q) is also a point on OK. Thus,

$q = \dfrac {y_1 + y_2}{x_1 + x_2}p$ ………. (2)

Solving (1) and (2) will give you $p = ± r \dfrac {x_1 + x_2}{\sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2}}$

The ‘±’ provides two sets of answers for (p of M) and (p’ for M’) as shown.

The corresponding values of q can be found via (2).

Selecting the correct M(p, q) is another story.

The point lies on both the circle and the perpendicular bisector of the segment connecting the points, so the midpoint of the (minor) arc is on the radius through the midpoint of the section.

By translation, we may assume that the circle (which has radius, say, $r$) is centered at $(0, 0)$. Now, the midpoint of that segment is $(x_0, y_0) = \frac{1}{2} (x_1 + x_2, y_1 + y_2)$, and so the midpoint along the arc is

$$\frac{r}{\sqrt{x_0^2 + y_0^2}} (x_0, y_0) \textrm{.}$$

Note that this formula fails when the midpoint is $(0, 0)$, which occurs precisely when the two red points on the circle are antipodal, i.e., when they are end points of a diameter.