3 point to circle. And get radius

Question

Me and my collegues stumbled upon a simple geometric question were non of us could provide a quick and descent answer. After some more puzzeling we still could not figure it out.

We all have the feeling it should not be that hard but fail at answering it. We hoped someone here could help us out!

Given 3 random points in a 2 dimensional space.

No point is the same position and the 3 points cannot be on the same line.

Question 1. Can a circle always be drawn which goes through these 3 points?

we assumed yes

Question 2. Is there a simple formula which gives you the radius of this circle?

Answer 1

The formula for circle at center $(a,b)$ with radius $r$ is: $(x -a)^2 + (y-b)^2 = r^2$.

Os if you have three points $(x_1,y_2), (x_2,y_2)$ and $(x_3,y_3)$ then they all belong to a circle and you know:

1: $(x_1 - a)^2 + (y_1-b)^2 = r^2$ and

2: $(x_2 - a)^2 + (y_2-b)^2 = r^2$ and

3: $(x_3 - a)^2 + (y_3-b)^2 = r^2$

Use those three equations and solve for $a,b,r$ using $x_i, y_i$ as constants.

I trust that you will be able to figure that out.

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Although it probably easier to find the equation of the line going through $A = (x_1,y_1)$ and $B = (x_2, y_2)$. [It is $y = m_1 x + b_1$. To find $m_1, b_1$ plug in $(x_1, y_1)$ and $(x_2, y_2)$ and see what you get.].

Take the midpoint $(x_m, y_m)$ of $A$ and $B$ and the find the equation of the perpendicular through the midpoint of $A,B$. [It is $y = -\frac 1{m_1} + b_m$. Plug in $x_m, y_m$ to solve for $b_m$].

Do that for the points $B$ and $C = (x_3, y_3)$ to get lint $y= m_2 + b_2$ and and the midpoint $(x_{m_2}, y_{m_2})$ so the perpendicular line $y= -\frac 1{m_2} + b_{m_2}$].

Find the intersection of the perpendicular lines at $(a,b)$ [by solving plugging $(a,b)$ into both $y = -\frac 1{m_1} + b_m$ and $y= -\frac 1{m_2} + b_{m_2}$]

That use geometry reasoning to conclude those lines intersect at the center of the circle.

The radius will $r = d(A,(a,b)) = d(B, (a,b)) = d(C, (a,b))$ wher $d((x_1, y_1), (a,b)) = \sqrt {(x_1 -a)^2 + (y_2-b)^2}$

Answer 2

Consider the coordinates of two points, relative to the first point

$$ \begin{array}{cc} x_2 = x_1 + \Delta x_2 & y_2 = y_1 + \Delta y_2 \\ x_3 = x_1 + \Delta x_3 & y_3 = y_1 + \Delta y_3 \end{array} $$

and the equation of the circle for the three points

$$ \begin{array}{c} (x_1-x_c)^2 + (y_1-y_c)^2 = r^2 \\ (x_1+\Delta x_2-x_c)^2 + (y_1 + \Delta y_2-y_c)^2 = r^2 \\ (x_1+\Delta x_3-x_c)^2 + (y_1 + \Delta y_3-y_c)^2 = r^2 \\ \end{array}$$

Now subtract the first equation from the second one, and the first equation from the third one to eliminate all quadratic terms of the center $x_c$ and $y_c$.

$$ \begin{array}{c} \Delta x_2^2 + 2 \Delta x_2 ( x_1-x_c) + \Delta y_2 (\Delta y_2 + 2 ( y_1-y_c)) = 0 \\ \Delta x_3^2 + 2 \Delta x_3 ( x_1-x_c) + \Delta y_3 (\Delta y_3 + 2 ( y_1-y_c)) = 0 \end{array}$$

Now solve these equations for the center of the circle ( $x_c$ and $y_c$)

$$ \begin{aligned} x_c & = x_1 + \frac{\Delta x_2^2 \Delta y_3 - \Delta x_3^2 \Delta y_2 + \Delta y_2^2 \Delta y_3 - \Delta y_3^2 \Delta y_2}{2 (\Delta x_2 \Delta y_3 - \Delta x_3 \Delta y_2)} \\ y_c & = y_1 + \frac{\Delta x_2^2 \Delta x_3 - \Delta x_3^2 \Delta x_2 + \Delta y_2^2 \Delta x_3 - \Delta y_3^2 \Delta x_2}{2 (\Delta x_3 \Delta y_2 - \Delta x_2 \Delta y_3)} \\ \end{aligned}$$

and use the first equation to solve for $r$

$$ \boxed{r = \sqrt{ \left(\tfrac{\Delta x_2^2 \Delta y_3 - \Delta x_3^2 \Delta y_2 + \Delta y_2^2 \Delta y_3 - \Delta y_3^2 \Delta y_2}{2 (\Delta x_2 \Delta y_3 - \Delta x_3 \Delta y_2)} \right)^2 + \left(\tfrac{\Delta x_2^2 \Delta x_3 - \Delta x_3^2 \Delta x_2 + \Delta y_2^2 \Delta x_3 - \Delta y_3^2 \Delta x_2}{2 (\Delta x_3 \Delta y_2 - \Delta x_2 \Delta y_3)} \right)^2 } } $$

This is the same procedure used in this answer to find the center and radius of a circle from three points, but analytically instead of numerically.