# 4.2 Circles

Recall that a *circle* in ^{2} is the set of points (*x*, *y*) that lie at a fixed distance, called the *radius*, from a fixed point, called the *centre* of the circle. We can use the techniques of coordinate geometry to find the equation of a circle with a given centre and radius.

Let a circle have centre *C*(*a*, *b*) and radius *r*. Then, if *P*(*x*, *y*) is an arbitrary point on the circumference of the circle, the distance *CP* equals *r*. It follows from the formula for the distance between two points in the plane that

## Theorem 2

The equation of a circle in ^{2} with centre (*a*, *b*) and radius *r* is

(*x* − *a*)^{2} + (*y* − *b*)^{2} = *r*^{2}.

For example, it follows from this formula that the circle with centre (−1, 2) and radius √3 has equation

which can be simplified to give

*x*^{2} + 2*x* + 1 + *y*_{2} − 4*y* + 4 = 3,

or

*x*^{2} + *y*^{2} + 2*x* − 4*y* + 2 = 0.

## Example 51

Determine the equation of each of the following circles, given the centre and radius:

**(a)**centre the origin, radius 1;**(b)**centre the origin, radius 4;**(c)**centre (3, 4), radius 2.

### Answer

We use the standard formula for the equation of a circle of given centre and radius given in Theorem 2.

**(a)**This circle has equation(

*x*− 0)^{2}+ (*y*− 0)^{2}= 1^{2},which can be rewritten in the form

*x*^{2}+*y*^{2}= 1.**(b)**This circle has equation(

*x*− 0)^{2}+ (*y*− 0)^{2}= 4^{2},which can be rewritten in the form

*x*^{2}+*y*^{2}= 16.**(c)**This circle has equation(

*x*− 3)^{2}+ (*y*− 4)^{2}= 2^{2},which can be rewritten in the form

*x*^{2}+*y*^{2}− 6*x*− 8*y*+ 21 = 0.

If we expand the brackets in equation (4.1) and collect the corresponding terms, we can rewrite equation (4.1) in the form

*x*^{2} + *y*^{2} − 2*ax* − 2*by* + (*a*^{2} + *b*^{2} − *r*^{2}) = 0.

Then, if we write *f* for −2*a*, *g* for −2*b* and *h* for *a*^{2} + *b*^{2} − *r*^{2}, this equation takes the form

*Note:* The coefficients of *x*^{2} and *y*^{2} are equal.

In many situations, equation (4.1) is more useful than equation (4.2) for determining the equation of a particular circle.

We have seen that the equation of a circle can be written in the form

On the other hand, given an equation of the form (4.2), can we determine whether it represents a circle? If it does represent a circle, can we determine its centre and radius?

For example, consider the set of points (*x*, *y*) in the plane that satisfy the equation

*Note:* In equation (4.3) the coefficients of *x*^{2} and *y*^{2} are both 1.

In order to transform equation (4.3) into an equation of the form equation (4.1), we use a technique called ‘completing the square’. We rewrite the terms that involve only *x* and the terms that involve only *y*, as follows:

*Note:* −2 is half the coefficient of *x* and +3 is half the coefficient of *y* in equation (4.3).

Substituting the expressions above into equation (4.3), we obtain

((*x* − 2)^{2} − 4) + ((*y* + 3)^{2} − 9) + 9 = 0,

that is,

(*x* − 2)^{2} + (*y* + 3)^{2} = 4.

*Note:* We can ‘read off’ the centre and radius of the circle from this equation.

It follows that equation (4.3) represents a circle with centre (2, −3) and radius 2.

In general, we can use the same method, ‘completing the square’, to rewrite the equation

*x*^{2} + *y*^{2} + *fx* + *gy* + ^{h} = 0

in the form

from which we can ‘read off’ the centre and radius.

*Note:* Again, the coefficients of *x*^{2} and *y*^{2} are both 1.

The constant in the first bracket is half the coefficient of *x* in the previous equation, and the constant in the second bracket is half the coefficient of *y*.

## Theorem 3

An equation of the form

*x*^{2} + *y*^{2} + *fx* + *gy* + ^{h} = 0

represents a circle with

if and only if .

*Remark:* It follows from equation (4.4) that if , then there are no points (*x*, *y*) that satisfy the equation *x*^{2} + *y*^{2} + *fx* + *gy* + *h* = 0; and if , then the given equation simply represents the single point .

## Example 52

Determine the condition on the numbers *f*, *g* and *h* in the equation

*x*^{2} + *y*^{2} + *fx* + *gy* + *h* = 0

for the circle with this equation to pass through the origin.

### Answer

Since the origin lies on the circle, its coordinates (0, 0) must satisfy the equation of the circle. Thus

0^{2} + 0^{2} + *f* × 0 + *g* × 0 + *h* = 0,

which reduces to the condition *h* = 0.

## Example 53

Determine the centre and radius of each of the circles given by the following equations.

**(a)***x*^{2}+*y*^{2}− 2*x*− 6*y*+ 1 = 0**(b)**3*x*^{2}+ 3*y*^{2}− 12*x*− 48*y*= 0

### Answer

**(a)**We can complete the square in the equation*x*^{2}+*y*^{2}− 2*x*− 6*y*+ 1 = 0to obtain

(

*x*− 1)^{2}− 1 + (*y*− 3)^{2}− 9 + 1 = 0,or

(

*x*− 1)^{2}+ (*y*− 3)^{2}= 9.So the circle has centre (1, 3) and radius √9 = 3.

(Alternatively, use the general formula for centre and radius in Theorem 3 with

*f*= −2,*g*= −6 and*h*= 1.)**(b)**Here the coefficients of*x*^{2}and*y*^{2}are both 3, so we divide the equation by 3 to obtainCompleting the square gives

(

*x*− 2)^{2}− 4 + (*y*− 8)^{2}− 64 = 0,or

(

*x*− 2)^{2}+ (*y*− 8)^{2}+ 68.So the circle has centre (2, 8) and radius

(Alternatively, apply the general formula for centre and radius to the equation in the form of equation (S.14), with

*f*= −4,*g*= −16 and*h*= 0.)

## Example 54

Determine the set of points (*x*, *y*) in ^{2} that satisfy each of the following equations.

**(a)***x*^{2}+*y*^{2}+*x*+*y*+ 1 = 0**(b)***x*^{2}+*y*^{2}− 2*x*+ 4*y*+ 5 = 0**(c)**2*x*^{2}+ 2*y*^{2}+*x*− 3*y*− 5 = 0

### Answer

**(a)**If we complete the square in the equationwe obtain the equation

This equation represents the empty set, since its left-hand side is always non-negative whereas its right-hand side is negative.

(Alternatively, we can use information about the quantity . In equation (S.15), we have

*f*= 1,*g*= 1 and*h*= 1. Thusso the set must be empty.)

**(b)**If we complete the square in the equationwe obtain the equation

(

*x*− 1)^{2}+ (*y*+ 2)^{2}= 0.Thus the set in the plane represented by equation (S.16) is the single point (1, −2).

(Alternatively, we can use information about the quantity . In equation (S.16), we have

*f*= −2,*g*= 4 and*h*= 5. Thusso the set is the single point

**(c)**Here the coefficients of*x*^{2}and*y*^{2}are both 2, so we divide the equation by 2 to giveIf we complete the square in equation (S.17), we obtain the equation

that is,

Thus the set in the plane represented by equation (S.17) is a circle with centre and radius .

(Alternatively, we can use information about the quantity . In equation (S.17), we have and . Thus

so the set is the circle with centre and radius