Well, I can answer it now
I found the answer...here
Lets say that the center of the circle is (x0, y0) and that the arc contains your two points (x1, y1) and (x2, y2). Then the radius is: r=sqrt((x1-x0)(x1-x0) + (y1-y0)(y1-y0)). So:
int r = (int)Math.sqrt((x1-x0)*(x1-x0) + (y1-y0)*(y1-y0));
int x = x0-r;
int y = y0-r;
int width = 2*r;
int height = 2*r;
int startAngle = (int) (180/Math.PI*atan2(y1-y0, x1-x0));
int endAngle = (int) (180/Math.PI*atan2(y2-y0, x2-x0));
graphics.drawArc(x, y, width, height, startAngle, endAngle);
Are the two red lines supposed to be perpendicular to one another, and at 45 degrees to the vertical, as shown? If so, and the circle center is at $(a, b)$ and the circle radius is $r$, then the points you're looking for are at
$$
(a + \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r) \\
(a - \frac{\sqrt{2}}{2} r, b + \frac{\sqrt{2}}{2}r).
$$
Since the two lines are (after the comments) not supposed to be perpendicular, let's call the angle between them $\alpha$; in your case, $\alpha = 85.702%{\circ}$. Let
$$\beta = 90^{\circ} - \frac{\alpha}{2};$$
Then $\beta$ is the half-angle between one of your dotted orange lines and the vertical greenish line. The offset of the two points to either side of that vertical line is therefore $r \sin \beta$, so we get
$$
(a + r \sin \beta, b + r \cos \beta) \\
(a - r \sin \beta, b + r \cos \beta).
$$
Applying these to your points, I get (using 3.14159 for PI, and Excel for the calculations)
right point = ( 0.073312432, -0.131990535)
left point = (-0.073312432, -0.131990535)
That's in pretty close agreement with your CAD results, and it's also mathematically correct. :)
Best Answer
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.