2 Engineering Example 1

2.1 A circle-cutting machine

Introduction

A cutting machine creates circular holes in a piece of sheet-metal by starting at the centre of the circle and cutting its way outwards until a hole of the correct radius exists. However, prior to cutting, the circle is characterised by three points on its circumference, rather than by its centre and radius. Therefore, it is necessary to be able to find the centre and radius of a circle given three points that it passes through.

Problem in words

Given three points on the circumference of a circle, find its centre and radius

1. for three general points
2. 1. for $\left(-6,5\right)$ , $\left(-3,6\right)$ and $\left(2,1\right)$
   2. for $\left(-0.7,0.6\right)$ , $\left(5.9,1.4\right)$ and $\left(0.8,-2.8\right)$

where coordinates are in cm.

Mathematical statement of problem

A circle passes through the three points. Find the centre $\left(x_0,y_0\right)$ and radius $R$ of this circle when the three circumferential points are

1. $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$
2. 1. $\left(-6,5\right)$ , $\left(-3,6\right)$ and $\left(2,1\right)$
   2. $\left(-0.7,0.6\right)$ , $\left(5.9,1.4\right)$ and $\left(0.8,-2.8\right)$

   Measurements are in centimetres; give answers correct to 2 decimal places.

Mathematical analysis

1. The equation of a circle with centre at $\left(x_0,y_0\right)$ and radius $R$ is

   ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=R^2$

   and, if this passes through the 3 points $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ then

   $$\begin{array}{rcll}{\left(x_1-x_0\right)}^2+{\left(y_1-y_0\right)}^2& =& R^2& \text{(1)}\\ {\left(x_2-x_0\right)}^2+{\left(y_2-y_0\right)}^2& =& R^2& \text{(2)}\\ {\left(x_3-x_0\right)}^2+{\left(y_3-y_0\right)}^2& =& R^2& \text{(3)}\end{array}$$

   Eliminating the $R^2$ term between (1) and (2) gives

   ${\left(x_1-x_0\right)}^2+{\left(y_1-y_0\right)}^2={\left(x_2-x_0\right)}^2+{\left(y_2-y_0\right)}^2$

   so that

   $x_1^2-2x_0{x}_1+y_1^2-2y_0{y}_1=x_2^2-2x_0{x}_2+y_2^2-2y_0{y}_2$ (4)

   Similarly, eliminating $R^2$ between (1) and (3) gives

   $x_1^2-2x_0{x}_1+y_1^2-2y_0{y}_1=x_3^2-2x_0{x}_3+y_3^2-2y_0{y}_3$ (5)

   Re-arranging (4) and (5) gives a system of two equations in $x_0$ and $y_0$ .

   $2\left(x_2-x_1\right)x_0+2\left(y_2-y_1\right)y_0=x_2^2+y_2^2-x_1^2-y_1^2$ (6)

   $2\left(x_3-x_1\right)x_0+2\left(y_3-y_1\right)y_0=x_3^2+y_3^2-x_1^2-y_1^2$ (7)

   Multiplying (6) by $\left(y_3-y_1\right)$ , and multiplying (7) by $\left(y_2-y_1\right)$ , subtracting and re-arranging gives

   $x_0=\frac{1}{2}\left(\frac{\left(y_3-y_1\right)\left(x_2^2+y_2^2\right)+\left(y_1-y_2\right)\left(x_3^2+y_3^2\right)+\left(y_2-y_3\right)\left(x_1^2+y_1^2\right)}{x_2{y}_3-x_3{y}_2+x_3{y}_1-x_1{y}_3+x_1{y}_2-x_2{y}_1}\right)$ (8)

   while a similar procedure gives

   $y_0=\frac{1}{2}\left(\frac{\left(x_1-x_3\right)\left(x_2^2+y_2^2\right)+\left(x_2-x_1\right)\left(x_3^2+y_3^2\right)+\left(x_3-x_2\right)\left(x_1^2+y_1^2\right)}{x_2{y}_3-x_3{y}_2+x_3{y}_1-x_1{y}_3+x_1{y}_2-x_2{y}_1}\right)$ (9)

   Knowing $x_0$ and $y_0$ , the radius $R$ can be found from

   $R=\sqrt{{\left(x_1-x_0\right)}^2+{\left(y_1-y_0\right)}^2}$ (10)

   (or alternatively using $x_2$ and $y_2$ (or $x_3$ and $y_3$ ) as appropriate).

   Equations (8), (9) and (10) can now be used to analyse the two particular circles above.

   1. Here $x_1=-6$ cm, $y_1=5$ cm, $x_2=-3$ cm, $y_2=6$ cm, $x_3=2$ cm and $y_3=1$ cm, so that

      $x_2{y}_3-x_3{y}_2+x_3{y}_1-x_1{y}_3+x_1{y}_2-x_2{y}_1=-3-12+10+6-36+15=-20$

      and

      $x_1^2+y_1^2=61x_2^2+y_2^2=45x_3^2+y_3^2=5$

      From (8)

      $x_0=\frac{1}{2}\left(\frac{-4\times 45+\left(-1\right)\times 5+5\times 61}{-20}\right)=\frac{-180-5+305}{-40}=-3$

      while (9) gives

      $y_0=\frac{1}{2}\left(\frac{-8\times 45+3\times 5+5\times 61}{-20}\right)=\frac{-360+15+305}{-40}=1$

      The radius can be found from (10)

      $R=\sqrt{{\left(-6-\left(-3\right)\right)}^2+{\left(5-1\right)}^2}=\sqrt{25}=5$

      so that the circle has centre at $\left(-3,1\right)$ and a radius of 5 cm.

   2. Now $x_1=-0.7$ cm, $y_1=0.6$ cm, $x_2=5.9$ cm, $y_2=1.4$ cm, $x_3=0.8$ cm and $y_3=-2.8$ cm, so that

      $x_2{y}_3-x_3{y}_2+x_3{y}_1-x_1{y}_3+x_1{y}_2-x_2{y}_1=-16.52-1.12+0.48-1.96-0.98-3.54=-23.64$

      and

      $x_1^2+y_1^2=0.85x_2^2+y_2^2=36.77x_3^2+y_3^2=8.48$

      so from (8)

      $x_0=\frac{1}{2}\left(\frac{-125.018-6.784+3.57}{-23.64}\right)=\frac{-128.232}{-47.28}=2.7121827$

      and from (9)

      $y_0=\frac{1}{2}\left(\frac{-55.155+55.968-4.335}{-23.64}\right)=\frac{-3.522}{-47.28}=0.0744924$

      and from (10)

      $R=\sqrt{{\left(-0.7-2.7121827\right)}^2+{\left(0.6-0.0744924\right)}^2}=\sqrt{11.9191490}=3.4524121$

      so that, to 2 d.p., the circle has centre at $\left(2.71,0.07\right)$ and a radius of 3.45 cm.

Mathematical comment

Note that the expression

$x_2{y}_3-x_3{y}_2+x_3{y}_1-x_1{y}_3+x_1{y}_2-x_2{y}_1$

appears in the denominator for both $x_0$ and $y_0$ . If this expression is equal to zero, the calculation will break down. Geometrically, this corresponds to the three points being in a straight line so that no circle can be drawn, or not all points being distinct so no unique circle is defined.