3.04 Graphs of circles

Determining the equation of a circle

The equation that defines a circle can be derived by directly applying Pythagoras' theorem.

Consider a circle of radius $r$r, centred at the origin of a Cartesian coordinate system. Any point $(x,y)$(x,y) on the circle must be $r$r units from the origin, by definition of a circle.

We can form a right-angled triangle by connecting the point $(x,y)$(x,y) to the point directly below (or above) it on the $x$x-axis, and the origin.

Applying Pythagoras' theorem to the side lengths of this triangle, we have that $x^2+y^2=r^2$x2+y2=r2. This relation defines the circle of radius $r$r, centred at the origin.

 

For circles centred away from the origin, the equation can be found using the same principles.

The right-angled triangle in this diagram has sides $x-h$x−h, $y-k$y−k and $r$r. Using Pythagoras' theorem again, we see that $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2. This equation defines the circle of radius $r$r, centred at the point $\left(h,k\right)$(h,k).

The equation could also be viewed as a translation of the circle centred at the origin, which has equation $x^2+y^2=r^2$x2+y2=r2, by $h$h units horizontally and $k$k units vertically.

 

If we expand each of the squared brackets in the equation that we have just derived, we can rewrite the relation in the form $x^2-2hx+h^2+y^2-2ky+k^2=r^2$x2−2hx+h2+y2−2ky+k2=r2.

By collecting the constant terms and relabelling the coefficients, we obtain an equation of the form $x^2+y^2+ax+by+c=0$x2+y2+ax+by+c=0. This is called the general form for an equation for a circle.

 

Finding the centre and radius from the equation

For the equation of a circle given in general form, how can we obtain information about the centre and the radius of the circle? It turns out that we can do this in one of two ways: equating coefficients or completing the square.

To equate coefficients, we look at the expanded form $x^2-2hx+h^2+y^2-2ky+k^2=r^2$x2−2hx+h2+y2−2ky+k2=r2 and compare it to the general form $x^2+y^2+ax+bx+c=0$x2+y2+ax+bx+c=0. We then equate coefficients to see that $a=-2h$a=−2h, $b=-2k$b=−2k and $c=h^2+k^2-r^2$c=h2+k2−r2.

So from the general form we have that $h=-\frac{a}{2}$h=−a2​, and similarly $k=-\frac{b}{2}$k=−b2​. Hence the centre of a circle given in general form will be the point $\left(-\frac{a}{2},-\frac{b}{2}\right)$(−a2​,−b2​). Once found, the values for $h$h and $k$k can then be used to find the radius by the formula $r=\sqrt{h^2+k^2-c}$r=√h2+k2−c.

 

Let's now look at some examples using the completing the square method. This is a useful method to have well practised in our toolbox and does not require memorisation of formula.

 

Worked examples

Example 1

Given the following expression for a circle in general form $x^2+2x-6y+y^2-6=0$x2+2x−6y+y2−6=0, find the radius of the circle and the location of its centre.

We complete the squares involving the $x$x- and $y$y-terms separately.

Halve the coefficient of the $x$x and $y$y term, square this and add it to both sides of the equation.

$(x^2+2x+1)+(y^2-6y+9)=6+1+9$(x2+2x+1)+(y2−6y+9)=6+1+9

$(x+1)^2+(y-3)^2=16$(x+1)2+(y−3)2=16, rewrite the terms grouped above as perfect squares.

$(x+1)^2+(y-3)^2=4^2$(x+1)2+(y−3)2=42, rewrite the right-hand side as a square.

Comparing this with the equations derived earlier, we conclude that the centre is at $(-1,3)$(−1,3) and the radius is $4$4.

Example 2

Determine whether the relation $x^2+6+y^2+5y=0$x2+6+y2+5y=0 defines a circle. If so, find its centre and radius.

Completing a square and moving the constants to the right, we have

$x^2+y^2+5y+\frac{25}{4}=-6+\frac{25}{4}$x2+y2+5y+254​=−6+254​

$x^2+(y+\frac{5}{2})^2=\frac{1}{4}$x2+(y+52​)2=14​

The terms on the left are in the required form and there is a positive number on the right. So, the relation does define a circle in the Cartesian plane. If the number on the right had been negative, the radius could not have been a real number.

We conclude that the centre is at $\left(0,-\frac{5}{2}\right)$(0,−52​) and the radius is $\sqrt{\frac{1}{4}}=\frac{1}{2}$√14​=12​.

 

Domain and range of circles

The diagram below shows that, for a circle of radius $r$r centred at the point $\left(h,k\right)$(h,k), the domain consists of all $x$x-values within the interval $h-r\le x\le h+r$h−r≤x≤h+r while the range consists of all $y$y-values within the interval $k-r\le y\le k+r$k−r≤y≤k+r.

Using interval notation, we can also write the domain as $\left[h-r,h+r\right]$[h−r,h+r], and the range as $\left[k-r,k+r\right]$[k−r,k+r].

 

Practice questions

Question 1

Question 2

Question 3

 

Semicircles

The circle itself is not a function, but can be split into two semicircles, each of which are functions. We can rearrange the equation to make this clear:

$\left(x-a\right)^2+\left(y-b\right)^2$(x−a)2+(y−b)2	$=$=	$r^2$r2
$\left(y-b\right)^2$(y−b)2	$=$=	$r^2-\left(x-a\right)^2$r2−(x−a)2
$y-b$y−b	$=$=	$\pm\sqrt{r^2-\left(x-a\right)^2}$±√r2−(x−a)2
$y$y	$=$=	$b\pm\sqrt{r^2-\left(x-a\right)^2}$b±√r2−(x−a)2

The positive square root gives the top half of the circle and the negative square root gives the lower half.

So, for example, a circle centred at the point $\left(2,5\right)$(2,5) with radius $r=3$r=3 can be written in the form $y=5\pm\sqrt{9-\left(x-2\right)^2}$y=5±√9−(x−2)2. This can then be split into two semicircular functions, say $f$f and $g$g, given by $f\left(x\right)=5+\sqrt{9-\left(x-2\right)^2}$f(x)=5+√9−(x−2)2 and $g\left(x\right)=5-\sqrt{9-\left(x-2\right)^2}$g(x)=5−√9−(x−2)2.

Each of these functions are semicircles. Both functions will have the same domain; that of the original circle, given by $-1\le x\le5$−1≤x≤5. The ranges are different, however. The range of $f$f is that of the top half of the circle, $5\le y\le8$5≤y≤8, while the range of $g$g is that of the bottom half of the circle, $2\le y\le5$2≤y≤5. This is show in the following image:

The circle can also be split into other pairs of semicircles, such as the left and right halves of the circle, but these semicircles will be relations and not functions. To find equations for the left and right halves, we follow a similar process to the above but by rearranging for $x$x instead. 

 

Practice questions

QUESTION 4

QUESTION 5