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AustraliaVIC
VCE 11 Methods 2023

4.08 Graphs of circles

Lesson

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.

Form a right-angled triangle by connecting the point $(x,y)$(x,y), 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 gives $x^2+y^2=r^2$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 $x-h$xh, $y-k$yk and $r$r. Using Pythagoras' theorem again, see that $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2. This equation defines the circle of radius $r$r, centred at the point $\left(h,k\right)$(h,k).

Equation of a circle in standard form

The equation of a circle in standard form is

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2,

where the circle has radius $r$r and is 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.

 

Expanding each of the squared brackets in the derived equation, the relation can be rewritten in the form $x^2-2hx+h^2+y^2-2ky+k^2=r^2$x22hx+h2+y22ky+k2=r2.

Collecting the constant terms and relabeling the coefficients gives 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.

Equation of a circle in general form

The standard form for the equation of a circle can be rearranged to give the general form of the equation of a circle:

$x^2+y^2+ax+by+c=0$x2+y2+ax+by+c=0

 

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 this can be done in one of two ways: completing the square or equating coefficients.

To equate coefficients, look at the expanded form $x^2-2hx+h^2+y^2-2ky+k^2=r^2$x22hx+h2+y22ky+k2=r2 and compare it to the general form $x^2+y^2+ax+bx+c=0$x2+y2+ax+bx+c=0. Next, equate coefficients to see that $a=-2h$a=2h, $b=-2k$b=2k and $c=h^2+k^2-r^2$c=h2+k2r2.

Rearranging gives $h=-\frac{a}{2}$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+k2c.

 

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+2x6y+y26=0, find the radius of the circle and the location of its centre.

Use the complete the square method for the $x$x- and $y$y-terms separately.

For each of the $x$x and $y$y terms, halve the coefficient, 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)+(y26y+9)=6+1+9

$(x+1)^2+(y-3)^2=16$(x+1)2+(y3)2=16, rewrite the terms grouped above as perfect squares.

$(x+1)^2+(y-3)^2=4^2$(x+1)2+(y3)2=42, rewrite the right-hand side as a square.

Comparing this with the equations derived earlier, the centre of the circle 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 $h-r\le x\le h+r$hrxh+r while the range consists of all $y$y-values within the interval $k-r\le y\le k+r$kryk+r.

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

 

Practice questions

Question 1

Consider the circle $x^2+y^2=4$x2+y2=4.

  1. Find the $x$x-intercepts. Write all solutions on the same line separated by a comma.

  2. Find the $y$y-intercepts. Write all solutions on the same line separated by a comma.

  3. Graph the circle.

    Loading Graph...

Question 2

Consider the equation $\left(x+5\right)^2+\left(y+3\right)^2=16$(x+5)2+(y+3)2=16.

  1. Plot the graph described by the equation.

    Loading Graph...

  2. State the domain of the graph, using interval notation.

  3. State the range of the graph, using interval notation.

Question 3

Consider the equation of a circle given by $y^2+2y+8=12x-x^2+7$y2+2y+8=12xx2+7.

  1. Rewrite the equation of the circle in the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2.

  2. What are the coordinates of the centre of this circle?

  3. What is the radius of this circle?

  4. Graph this circle on the axes below:

    Loading Graph...

 

Semicircles

The circle itself is not a function, but can be split into two semicircles, each of which are functions. Rearranging the equations makes this clear:

$\left(x-a\right)^2+\left(y-b\right)^2$(xa)2+(yb)2 $=$= $r^2$r2
$\left(y-b\right)^2$(yb)2 $=$= $r^2-\left(x-a\right)^2$r2(xa)2
$y-b$yb $=$= $\pm\sqrt{r^2-\left(x-a\right)^2}$±r2(xa)2
$y$y $=$= $b\pm\sqrt{r^2-\left(x-a\right)^2}$b±r2(xa)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(x2)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(x2)2 and $g\left(x\right)=5-\sqrt{9-\left(x-2\right)^2}$g(x)=59(x2)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$1x5. The ranges are different, however. The range of $f$f is that of the top half of the circle, $5\le y\le8$5y8, while the range of $g$g is that of the bottom half of the circle, $2\le y\le5$2y5. This is shown 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, follow a similar process to the above but by rearranging for $x$x instead. 

 

Practice questions

QUESTION 4

Answer the following.

  1. Rearrange the equation $x^2+y^2=4$x2+y2=4 to make $y$y the subject.

  2. Write down the equation of the semicircle with $y$y as the subject which has a radius $2$2 and non-positive $y$y-values.

  3. Draw the graph of the semicircle below.

    Loading Graph...

QUESTION 5

The top of a semicircle has a domain of $\left[-10,2\right]$[10,2] and a range of $\left[-2,4\right]$[2,4].

  1. Plot the semicircle.

    Loading Graph...

  2. State the equation for the semicircle in the form $y=\pm\sqrt{r^2-\left(x-h\right)^2}+k$y=±r2(xh)2+k.

Outcomes

U1.AoS1.2

qualitative interpretation of features of graphs of functions, including those of real data not explicitly represented by a rule, with approximate location of any intercepts, stationary points and points of inflection

U1.AoS1.12

draw graphs of polynomial functions of low degree, simple power functions and simple relations that are not functions

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