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
- for three general points
-
- for [maths rendering] , [maths rendering] and [maths rendering]
- for [maths rendering] , [maths rendering] and [maths rendering]
where coordinates are in cm.
Mathematical statement of problem
A circle passes through the three points. Find the centre [maths rendering] and radius [maths rendering] of this circle when the three circumferential points are
- [maths rendering] , [maths rendering] and [maths rendering]
-
- [maths rendering] , [maths rendering] and [maths rendering]
- [maths rendering] , [maths rendering] and [maths rendering]
Measurements are in centimetres; give answers correct to 2 decimal places.
Mathematical analysis
-
The equation of a circle with centre at
[maths rendering]
and radius
[maths rendering]
is
[maths rendering]
and, if this passes through the 3 points [maths rendering] , [maths rendering] and [maths rendering] then
[maths rendering]Eliminating the [maths rendering] term between (1) and (2) gives
[maths rendering]
so that
[maths rendering] (4)
Similarly, eliminating [maths rendering] between (1) and (3) gives
[maths rendering] (5)
Re-arranging (4) and (5) gives a system of two equations in [maths rendering] and [maths rendering] .
[maths rendering] (6)
[maths rendering] (7)
Multiplying (6) by [maths rendering] , and multiplying (7) by [maths rendering] , subtracting and re-arranging gives
[maths rendering] (8)
while a similar procedure gives
[maths rendering] (9)
Knowing [maths rendering] and [maths rendering] , the radius [maths rendering] can be found from
[maths rendering] (10)
(or alternatively using [maths rendering] and [maths rendering] (or [maths rendering] and [maths rendering] ) as appropriate).
Equations (8), (9) and (10) can now be used to analyse the two particular circles above.
-
Here
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm and
[maths rendering]
cm, so that
[maths rendering]
and
[maths rendering]
From (8)
[maths rendering]
while (9) gives
[maths rendering]
The radius can be found from (10)
[maths rendering]
so that the circle has centre at [maths rendering] and a radius of 5 cm.
-
Now
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm and
[maths rendering]
cm, so that
[maths rendering]
and
[maths rendering]
so from (8)
[maths rendering]
and from (9)
[maths rendering]
and from (10)
[maths rendering]
so that, to 2 d.p., the circle has centre at [maths rendering] and a radius of 3.45 cm.
-
Here
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm,
[maths rendering]
cm and
[maths rendering]
cm, so that
Mathematical comment
Note that the expression
[maths rendering]
appears in the denominator for both [maths rendering] and [maths rendering] . 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.