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I want to place a new point in the middle of the two points which are on the circle outline (Arc). I have the coordinates $(x,y)$ of the center of the circle, the two red points and the radius of the circle. I want to find out the coordinates of the midpoint of the Arc. Is there a formula for this ?

Example: I want the midpoint coordinates between the two red lines.
circle with two points on the line

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Express the coordinates in polar form, then the midpoints coordinates are at their average. – Macavity Sep 4 '14 at 14:20
You would also need the coordinates of the center of the circle. Also maybe re-word it -- "the two red points" not "the two red lines"... – coffeemath Sep 4 '14 at 14:20
As stated, no, as the midpoint is not even well-defined: reflecting the circle of your drawing in the line between the two points, you get another circle with the same radius, but the midpoint is no longer the same. – fuglede Sep 4 '14 at 14:27
@Macavity, polar representations are not unique, and different choices here will give different midpoints! (Sometimes that of the minor arc, sometimes that of the major.) – Travis Sep 4 '14 at 14:29
@fuglede From context ("on the circle outline") it seems that the circle itself is fixed; if not, you are of course correct. – Travis Sep 4 '14 at 14:32
up vote 3 down vote accepted

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.

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thanks this did the job. Just the formula is too big – Devid Sep 4 '14 at 17:08
@Devid You are welcome. Also, I notice that you have edited my post and remarked that making the diagram clickable. Would like to know was the diagram uploaded not clickable in the original version? If it is a yes, do let me know what else I should do before uploading it. – Mick Sep 5 '14 at 3:46
yes when you upload it the normal way the diagram is not clickable. To make it clickable put the image url and put it like this: <a href="image.png"> <img src="image.png"> </a> Now when you hold Ctrl and click on the image it will open in a new tab. – Devid Sep 5 '14 at 9:59
@Devid Thanks and glad to learn something new. – Mick Sep 5 '14 at 15:39

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.

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You can make a system of two equations and solve for the coordinate values of your new point. One equation is that the distance between your new point and the center equals the radius. (So, you need to know the formula for distance). The other equation is that the distance between your new point and one red point equals the distance between your new point and the other red point.

Setting these equations up and solving them will give you two possible points on opposite sides of the circle. The one closest to your two red points is probably the one you want.

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The line perpendicular to $AB$ through $O$ gives you the two solutions. If $O=(0,0)$ (for simplicity), $A=(x_A,y_A)$, $B=(x_B,y_B)$, this line is given by $(x,y)=(t\cdot(y_B-y_A), t\cdot (x_A-x_B))$. You need to pick $t$ such that the distance from $O$ becomes the radius $r$, i.e. $t=\pm\frac{r}{\sqrt{(y_B-y_A)^2+(x_B-x_A)^2}}$.

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