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.