Hello, strewart!

I have an approach, but I did not finish the problem . . .

As part of an experiment I am doing, I need a radius for a circle.

I have the lengths of three segments connecting the three points on the circle.

Is there a way to work out the radius of the circle from that?

I know with two points, the circle could be any size, .Right!

but with 3 points there should only be one solution for a circle through all three. .Yes!

I've tried various simultaneous equations and not been able to come up with a solution.

I always end up with 2 unknowns in any equation I make.

Any help would be appreciated. Code:

C
*
b * * a
* * *
* *
A * - - - - * - - - - * B
* *
r * β * α * r
* *
*
O

We have $\displaystyle \Delta ABC$ on a circle with center $\displaystyle O$ and radius $\displaystyle r.$

. . $\displaystyle r \,=\,AO \,=\,BO \,=\,CO.$

$\displaystyle a \,=\,BC,\;b\,=\,AC,\;c\,=\,AB.$

Let $\displaystyle \alpha \,=\,\angle BOC,\; \beta \,=\,\angle AOC$

In $\displaystyle \Delta BOC$, Law of Cosines:

. . $\displaystyle a^2 \;=\;r^2 + r^2 - 2r^2\cos\alpha \;=\; 2r^2(1-\cos\alpha) \;=\;4r^2\left(\frac{1-\cos\alpha}{2}\right)$

. . $\displaystyle a^2 \;=\;4r^2\sin^2\tfrac{\alpha}{2} \quad\Rightarrow\quad a \;=\;2r\sin\tfrac{\alpha}{2} $

Hence, we have: .$\displaystyle \begin{Bmatrix}a &=& 2r\sin\frac{\alpha}{2} & [1] \\ \\[-3mm] b &=& 2r\sin\frac{\beta}{2} & [2] \\ \\[-3mm] c &=& 2r\sin\frac{\alpha+\beta}{2} & [3] \end{Bmatrix}$

$\displaystyle \text{From [1]: }\:\sin\tfrac{\alpha}{2} \:=\:\frac{a}{2r} \quad\Rightarrow\quad \cos\tfrac{\alpha}{2} \:=\:\frac{\sqrt{4r^2-a^2}}{2r}\;\;[4] $

$\displaystyle \text{From [2]: }\:\sin\tfrac{\beta}{2} \:=\:\frac{b}{2r} \quad\Rightarrow\quad \cos\tfrac{\beta}{2} \:=\:\frac{\sqrt{4r^2-b^2}}{2r}\;\;[5] $

$\displaystyle \text{From [3]: }\:\sin\tfrac{\alpha+\beta}{2} \:=\:\frac{c}{2r}$

. . $\displaystyle \sin\tfrac{\alpha}{2}\cos\tfrac{\beta}{2} + \cos\tfrac{\alpha}{2}\sin\tfrac{\beta}{2} \;=\;\frac{c}{2r}$

Substitute [4] and [5]:

. . $\displaystyle \frac{a}{2r}\!\cdot\!\frac{\sqrt{4r^2-b^2}}{2r} + \frac{\sqrt{4r^2-a^2}}{2r}\!\cdot\!\frac{b}{2r} \;=\;\frac{c}{2r}$

. . . . . . $\displaystyle a\sqrt{4r^2-b^2} + b\sqrt{4r^2-a^2} \;=\;2cr$

And now, *all we have to do* is: eliminate the radicals and solve for *r.*