Calculating radius of a circle from 3 arcs

**strewart** · June 5th 2011, 09:10 PM

Hey all, as part of an experiment I am doing I need a radius for a circle, but the problem is I can't see all of it so I can't just measure the circle. Thus the information I have is the lengths of three connecting points on the outside of the circle, as I have shown in the diagram attached. 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, but with 3 points there should only be one solution for a circle touching all three. I have tried various simultaneous equations and not been able to come up with a solution for it, I always end up with 2 unknowns in any equation I make.

Any help would be appreciated.

Edit: Apparently it is not a valid image file for some reason.. Here is a link:

http://i218.photobucket.com/albums/c...art/circle.jpg

**Prove It** · June 5th 2011, 09:25 PM

Do you know the coordinates of the three vertices of your triangle? If so, just substitute them into

$\displaystyle (x - h)^2 + (y - k)^2 = r^2$

which will give you three equations in three unknowns to solve for.

**strewart** · June 5th 2011, 11:44 PM

I don't have coordinates, no. Just the length of the three chords. It did occur to me that it would probably be a lot easier to work out using coordinates instead; the line perpendicular to the chord at its midpoint will go through the centre of the circle, find out where two of them intersect and its easy enough. Is it possible to do it just using the lengths of the chords though? I'd have to set it up on graph paper and define an x and y axis to do it with coordinates and I want to get it as accurate as possible.

**Soroban** · June 6th 2011, 05:56 AM

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 $\Delta ABC$ on a circle with center $O$ and radius $r.$
. . $r \,=\,AO \,=\,BO \,=\,CO.$

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

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

In $\Delta BOC$ , Law of Cosines:

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

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

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

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

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

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

Substitute [4] and [5]:

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

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

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

**bjhopper** · June 6th 2011, 06:23 AM

Hi Steward,
Use coordinate geometry. Place the longest side along the x axis starting at (0,0)
Next at (0,0) write the equation of a circle with radius of another side. Do the same
at the other end for the last side. You will have two equations in x and y and lastly a quadratic equation in one unknown.This will provide you with the three vertices of an upper or lower congruent triangle.Take it from there.

bjh

**Croat** · June 6th 2011, 06:56 AM

Hey all, as part of an experiment I am doing I need a radius for a circle, but the problem is I can't see all of it so I can't just measure the circle. Thus the information I have is the lengths of three connecting points on the outside of the circle, as I have shown in the diagram attached. 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, but with 3 points there should only be one solution for a circle touching all three. I have tried various simultaneous equations and not been able to come up with a solution for it, I always end up with 2 unknowns in any equation I make.

So if I get it right, you have a triangle and the size of it's sides?
If so then just use Heron's formula for calculating the area of the triangle and then use the alternative formula for the Area ( $A=\frac{abc}{4R}$ ) where R is the radius of the circumscribed circle, to get the Radius of your circle.

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