How to get enough data to draw an arc or a curve from 3 points?

Question

Note: Although I want to accomplish this in java, I think the question is more suitable for this site since it is mostly mathematical.

I am in the following scenario. I want to draw a curve and I have the following points:

• The starting point
• The midpoint of the curve
• The end point

And java allows me to create a curve from any of the following:

• an arc with 2 points, the starting angle and the ending angle
• an arc with center, radius, start angle and end angle
• an arc with 3 points and the radius
• Other curves of different types with 1 or 2 "control points", which I see that I do not have.

This Gimp screenshot shows roughly what I'm trying to achieve.

Edit: I clarify that I want to achieve the type of curves or arcs used in perspective to create circles. Next I leave a photo (it does not belong to me) to illustrate:

Edit 2:

One pattern I noticed, is that the curve is "enclosed" in a square formed by the first and last points. Below is a picture of an "ideal" curve:

What formula can I use to calculate any of those things?

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Doing research, I found several questions that didn't serve my purpose (both here and on Stack Overflow) or I wasn't able to understand. Among what I found, I saw that you can get the center of an arc with bisectors as it says here, but I don't feel comfortable working with the long equations of the rect.

Answer 1

One way you can solve this problem is by using Bezier quadratic curves. The curve is given parametrically by

$ P(t) = P_1 (1 - t)^2 + 2 t (1 - t) P_2 + t^2 P_3 $

where $P_1$ is the starting point, $P_3$ is the end point, and $P_2$ is the point where the tangents to this (parabola) at $P_1$ and $P_3$ intersect.

You have $P_1$ and $P_3$ but you don't have $P_2$. Instead you have the midpoint of the curve at some $t \in [0, 1] $. Let's call this point $P_4$, so what you have is $P_1$ (start), $P_3$ (end) and $P_4$ (midpoint). What you don't have is $P_3$ and the value of $t = t_0$ at which $P(t_0) = P_4$.

Let your parameter be $t_0$, then

$ P(t_0) = P_1 (1 - t_0)^2 + 2 t_0 (1 - t_0) P_2 + t_0^2 P_3 = P_4 $

For example, we can take $t_0$ to be equal to $0.5$. Substituting this, we get

$ P_2 = \dfrac{ P_4 - P_1 (1 - t_0)^2 - t_0^2 P_3 }{2 t_0 (1 - t_0) } $

Since we've selected $t_0 = 0.5$ then

$ P_2 = \dfrac{ P_4 - 0.25 P_1 - 0.25 P_3}{0.5} = 2 P_4 - 0.5 P_1 - 0.5 P_3 $

Now that we have $P_2$, we can draw the curve $P(t)$ from $t = 0$ to $t = 1$.

Below are two examples of this method. The red points are $P_1$ through $P_4$ with $P_2$ being the calculated point and lies outside the curve while $P_4$ is the given midpoint.

Answer 2

Note: this answer is just passing the parametric equations from another answer into equations to get x and y separately. If this answer helped you, don't forget to support whoever did the original reasoning.

I also apologize for not using latex, since I don't know how to use it on this site, any editing or guidance is welcome.

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We have the following parametric equation:

P2 = 2P4 − 0.5P1 − 0.5P3

We make explicit the points involved and resolve

(x, y) = 2(bx, by) - 0.5(ax, ay) - 0.5(cx, cy)
(x, y) = (2bx, 2by) - (0.5ax, 0.5ay) - (0.5cx, 0.5cy)
(x, y) = (2bx - 0.5ax - 0.5cx, 2by - 0.5ay - 0.5cy)

What results in, being k the axis with which the calculations are made

k = 2bk - 0.5ak - 0.5ck