1 Equations for circles in the $Oxy$ plane

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 $R$ or $r$ or $a$ .

In coordinate geometry terms suppose $\left(x,y\right)$ denotes the coordinates of a point. For example, (4,2) means $x=4,\phantom{\rule{1em}{0ex}}y=2$ , $\left(-1,1\right)$ means $x=-1,\phantom{\rule{1em}{0ex}}y=1$ and so on. See Figure 26.

Figure 26

Example 14

Write down the distances ${d}_{1}$ and ${d}_{2}$ from the origin of the points with coordinates (4,2) and $\left(-1,1\right)$ respectively. Generalise the result to obtain the distance $d$ from the origin of any arbitrary point with coordinates $\left(x,y\right)$ .

Solution

Using Pythagoras’ Theorem:

${d}_{1}=\sqrt{{4}^{2}+{2}^{2}}=\sqrt{20}$ is the distance between the origin (0,0) and the point (4,2).

${d}_{2}=\sqrt{{\left(-1\right)}^{2}+{1}^{2}}=\sqrt{2}$ is the distance between the origin and $\left(-1,1\right)$ .

$d=\sqrt{{x}^{2}+{y}^{2}}$ is the distance from the origin to an arbitrary point $\left(x,y\right)$ . Note that the positive square root is taken in each case.

1.1 Circles with centre at the origin

Suppose $\left(x,y\right)$ is any point $P$ on a circle of radius $R$ whose centre is at the origin. See Figure 27.

Figure 27

Using the final result of Example (14), write down an equation relating $x,\phantom{\rule{1em}{0ex}}y$ and $R$ .

Since $\sqrt{{x}^{2}+{y}^{2}}$ is distance of any point $\left(x,y\right)$ from the origin, then for any point $P$ on the above circle.

$\phantom{\rule{2em}{0ex}}\sqrt{{x}^{2}+{y}^{2}}=R\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}={R}^{2}$

As the point $P$ in Figure 27 moves around the circle its $x$ and $y$ coordinates change. However $P$ will remain at the same distance $R$ from the origin by the very definition of a circle.

Hence we say that

$\phantom{\rule{2em}{0ex}}\sqrt{{x}^{2}+{y}^{2}}=R$ or, more usually,

${x}^{2}+{y}^{2}={R}^{2}$ (1)

is the equation of the circle radius $R$ centre at the origin. What this means is that if a point $\left(x,y\right)$ satisfies (1) then it lies on the circumference of the circle radius $R$ . If $\left(x,y\right)$ 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.

1. Write down the equation of this circle.
2. For the following points determine which lie on the circumference of this circle, which lie inside the circle and which lie outside the circle.

$\left(5,0\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(0,-5\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(4,3\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(-3,4\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(2,\sqrt{21}\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(-2\sqrt{6},1\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(1,4\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(4,-4\right)$

1. ${x}^{2}+{y}^{2}={5}^{2}=25$ is the equation of the circle.
2. For each point $\left(x,y\right)$ we calculate ${x}^{2}+{y}^{2}$ . If this equals 25 the point lies on the circle’ if greater than 25 then outside and if less than 25 then inside.

 $x,y$ ${x}^{2}+{y}^{2}$ conclusion $\left(5,0\right)$ 25 on circle $\left(0,-5\right)$ 25 on circle $\left(4,3\right)$ 25 on circle $\left(-3,-4\right)$ 25 on circle $\left(2,\sqrt{21}\right)$ 25 on circle $\left(-2\sqrt{6},1\right)$ 25 on circle $\left(1,4\right)$ 17 inside circle $\left(4,-4\right)$ 32 outside circle

Figure 28 demonstrates some of the results of the previous Task.

Figure 28

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 ${P}_{1}\left(-1,1\right)$ and ${P}_{2}\left(4,2\right)$ .

Using Pythagoras’ Theorem the distance between the two given points is

$\phantom{\rule{2em}{0ex}}d=\sqrt{{\left({P}_{1}A\right)}^{2}+{\left(A{P}_{2}\right)}^{2}}$

where ${P}_{1}A=4-\left(-1\right)=5,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}$ $A{P}_{2}=2-1=1$

$\therefore \phantom{\rule{1em}{0ex}}d=\sqrt{{5}^{2}+{1}^{2}}=\sqrt{26}$

Generalise your result to the previous Task to obtain the distance between any two points whose coordinates are $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ .

Between the arbitrary points ${P}_{1}$ and ${P}_{2}$ the distance is

$\phantom{\rule{2em}{0ex}}d=\sqrt{{\left(A{P}_{2}\right)}^{2}+{\left({P}_{1}A\right)}^{2}}$

where $A{P}_{2}={x}_{2}-{x}_{1}$ , ${P}_{1}A={y}_{1}-{y}_{2}=-\left({y}_{2}-{y}_{1}\right)$

so $d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Using the result of the last Task, we now consider a circle centre at the point $C\left({x}_{0};{y}_{0}\right)$ and of radius $R$ . Suppose $P$ is an arbitrary point on this circle which has co-ordinates $\left(x,y\right)$ :

Figure 29

Clearly $R=CP=\sqrt{{\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}}$

Hence, squaring both sides,

${\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}={R}^{2}$ (2)

which is said to be the equation of the circle centre $\left({x}_{0},{y}_{0}\right)$ radius $R$ .

Note that if ${x}_{0}={y}_{0}=0$ (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 $\left(x,y\right)$ satisfying (2) lies on the circumference of the circle.

Example 15

What does the equation $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}=4$   represent?

Solution

It represents a circle of radius 2 (the positive square root of 4) and has centre $C$ $\left(3,4\right)$ .

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 ( $\sqrt{4}$ ) and centre $\left(3,4\right)$ of the circle.

Write down the equations of each of the following circles for which the centre $C$ and radius $R$ are given:

1. $C\left(0,2\right),\phantom{\rule{1em}{0ex}}R=2$
2. $C\left(-2,0\right),\phantom{\rule{1em}{0ex}}R=3$
3. $C\left(-3,4\right),\phantom{\rule{1em}{0ex}}R=5$
4. $C\left(1,1\right),\phantom{\rule{1em}{0ex}}R=\sqrt{3}$
1. ${x}_{0}=0,\phantom{\rule{1em}{0ex}}{y}_{0}=2,\phantom{\rule{1em}{0ex}}{R}^{2}=4$ so by Equation (2) the circle’s equation is

$\phantom{\rule{2em}{0ex}}{x}^{2}+{\left(y-2\right)}^{2}=4$

2. ${x}_{0}=-2,\phantom{\rule{1em}{0ex}}{y}_{0}=0,\phantom{\rule{1em}{0ex}}{R}^{2}=9\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}{\left(x+2\right)}^{2}+{y}^{2}=9$
3. ${\left(x+3\right)}^{2}+{\left(y-4\right)}^{2}=25$
4. ${\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}=3$

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:

1. ${x}^{2}+{y}^{2}=4$ (centre $\left(0,0\right)$ i.e. the origin, radius 2)
2. ${\left(x-1\right)}^{2}+{y}^{2}=1$ (centre $\left(1,0\right)$ , radius 1)
3. ${\left(x-3\right)}^{2}+{\left(y-3\right)}^{2}=0$ (centre $\left(3,3\right)$ , radius 3)
4. ${\left(x+1\right)}^{2}+{\left(y+1\right)}^{2}=1$ (centre $\left(-1,-1\right)$ , radius 1)

Consider again the equation of the circle, centre (3,4) of radius 2:

$\phantom{\rule{2em}{0ex}}{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}=4$ (3)

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

$\phantom{\rule{2em}{0ex}}{x}^{2}-6x+9+{y}^{2}-8x+16=4$ or $\phantom{\rule{1em}{0ex}}{x}^{2}-6x+{y}^{2}-8x+21=0$ (4)

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 $R$

$\phantom{\rule{2em}{0ex}}{\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}={R}^{2}$

expand out the square terms and simplify.

We obtain

$\phantom{\rule{2em}{0ex}}{x}^{2}-2{x}_{0}x+{x}_{0}^{2}+{y}^{2}-2{y}_{0}y+{y}_{0}^{2}-{R}^{2}=0$

or

$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}-2{x}_{0}x-2{y}_{0}y+c=0$

where the constant $c={x}_{0}^{2}+{y}_{0}^{2}-{R}^{2}$ .

It follows from the above task that any equation of the form

$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}-2gx-2fy+c=0$ (5)

represents a circle with centre $\left(g,f\right)$ and a radius obtained by solving

$\phantom{\rule{2em}{0ex}}c={g}^{2}+{f}^{2}-{R}^{2}$

for $R$ .

Thus

$\phantom{\rule{2em}{0ex}}R=\sqrt{{g}^{2}+{f}^{2}-c}$ (6)

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. ${\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}={R}^{2}$ ) and hence obtain the centre and radius of the circle.

NB. The key point about Equation (5) is that the coefficients of the term ${x}^{2}$ and ${y}^{2}$ are the same , i.e. 1. An equation with the coefficient of ${x}^{2}$ and ${y}^{2}$ identical with value $k\ne 1$ could be converted into the form (5) by division of the whole equation by $k$ .

If

$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}-2x+10y+16=0$

obtain the centre and radius of the circle that this equation represents.

Begin by completing the square separately on the $x-$ terms and the $y-$ terms:

$\begin{array}{rcll}{x}^{2}-2x& =& {\left(x-1\right)}^{2}-1& \text{}\\ {y}^{2}+10y& =& {\left(y+5\right)}^{2}-25& \text{}\end{array}$

Now complete the problem:

The original equation

$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}-2x+10y+16=0$

becomes

$\phantom{\rule{2em}{0ex}}{\left(x-1\right)}^{2}-1+{\left(y+5\right)}^{2}-25+16=0$

$\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}{\left(x-1\right)}^{2}+{\left(y+5\right)}^{2}=10$

which represents a circle with centre $\left(1,-5\right)$ and radius $\sqrt{10}$ .

1.2 Circles and functions

$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}=1$

Solving for $y$ we obtain

$\phantom{\rule{2em}{0ex}}y=±\sqrt{1-{x}^{2}}$ .

This equation does not represent a function because of the two possible square roots which imply that for any value of $x$ there are two values of $y$ . (You will recall from earlier in this Workbook that a function requires only one value of the dependent variable $y$ corresponding to each value of the independent variable $x$ .)

However two functions can be obtained in this case: $y={y}_{1}=+\sqrt{1-{x}^{2}}\phantom{\rule{2em}{0ex}}y={y}_{2}=-\sqrt{1-{x}^{2}}$

whose graphs are the semicircles shown.

Figure 30