The obvious characteristic of a circle is that every point on its circumference is the same distance from the centre . This fixed distance is called the radius of the circle and is generally denoted by or or .
In coordinate geometry terms suppose denotes the coordinates of a point. For example, (4,2) means , means and so on. See Figure 26.
Write down the distances and from the origin of the points with coordinates (4,2) and respectively. Generalise the result to obtain the distance from the origin of any arbitrary point with coordinates .
Using Pythagoras’ Theorem:
is the distance between the origin (0,0) and the point (4,2).
is the distance between the origin and .
is the distance from the origin to an arbitrary point . Note that the positive square root is taken in each case.
Suppose is any point on a circle of radius whose centre is at the origin. See Figure 27.
Using the final result of Example (14), write down an equation relating and .
Since is distance of any point from the origin, then for any point on the above circle.
As the point in Figure 27 moves around the circle its and coordinates change. However will remain at the same distance from the origin by the very definition of a circle.
Hence we say that
or, more usually,
is the equation of the circle radius centre at the origin. What this means is that if a point satisfies (1) then it lies on the circumference of the circle radius . If does not satisfy (1) then it does not lie on that circumference.
Note carefully that the right-hand sides of the circle equation (1) is the square of the radius.
Consider the circle centre at the origin and of radius 5.
- Write down the equation of this circle.
For the following points determine which lie on the circumference of this circle,
which lie inside the circle and which lie outside the circle.
- is the equation of the circle.
- For each point we calculate . If this equals 25 the point lies on the circle’ if greater than 25 then outside and if less than 25 then inside.
Figure 28 demonstrates some of the results of the previous Task.
Note that the circle centre at the origin and of radius 1 has a special name – the unit circle.
Calculate the distance between the points and .
Using Pythagoras’ Theorem the distance between the two given points is
Generalise your result to the previous Task to obtain the distance between any two points whose coordinates are and .
Between the arbitrary points and the distance is
Using the result of the last Task, we now consider a circle centre at the point and of radius . Suppose is an arbitrary point on this circle which has co-ordinates :
Hence, squaring both sides,
which is said to be the equation of the circle centre radius .
Note that if (i.e. circle centre is at origin) then (2) reduces to (1) so the latter is simply a special case.
The interpretation of (2) is similar to that of (1): any point satisfying (2) lies on the circumference of the circle.
What does the equation represent?
It represents a circle of radius 2 (the positive square root of 4) and has centre .
N.B. There is no need to expand the terms on the left-hand side of the equation here. The given form reveals quite plainly the radius ( ) and centre of the circle.
Write down the equations of each of the following circles for which the centre and radius are given:
so by Equation (2) the circle’s equation is
Again we emphasise that the right-hand side of each of these equations is the square of the radius.
Write down the equations of each of the circles shown below:
- (centre i.e. the origin, radius 2)
- (centre , radius 1)
- (centre , radius 3)
- (centre , radius 1)
Consider again the equation of the circle, centre (3,4) of radius 2:
In this form of the equation the centre and radius of the circle can be clearly identified and, as we said, there is no advantage in squaring out. However, if we did square out the equation would become
Equation (4) is of course a valid equation for this circle but, we cannot immediately obtain the centre and radius from it.
For the case of the general circle of radius
expand out the square terms and simplify.
where the constant .
It follows from the above task that any equation of the form
represents a circle with centre and a radius obtained by solving
There is no need to remember Equation (6). In any specific problem the technique of completion of square can be used to turn an equation of the form (5) into the form of Equation (2) (i.e. ) and hence obtain the centre and radius of the circle.
NB. The key point about Equation (5) is that the coefficients of the term and are the same , i.e. 1. An equation with the coefficient of and identical with value could be converted into the form (5) by division of the whole equation by .
obtain the centre and radius of the circle that this equation represents.
Begin by completing the square separately on the terms and the terms:
Now complete the problem:
The original equation
which represents a circle with centre and radius .
Let us return to the equation of the unit circle
Solving for we obtain
This equation does not represent a function because of the two possible square roots which imply that for any value of there are two values of . (You will recall from earlier in this Workbook that a function requires only one value of the dependent variable corresponding to each value of the independent variable .)
However two functions can be obtained in this case:
whose graphs are the semicircles shown.