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3.07 Circles and semi-circles

Lesson

The equation of a circle with centre at the origin

The equations for a circle can be derived using Pythagoras' theorem. To do this we will consider the relatively simple case of a circle, 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.

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

In this triangle, we have by Pythagoras that $x^2+y^2=r^2$x2+y2=r2 and this relation defines the circle.

General equation for circle with centre at the origin

$x^2+y^2=r^2$x2+y2=r2

where $\left(x,y\right)$(x,y) are a pair of coordinates. The centre is $\left(0,0\right)$(0,0) and $r$r is the radius

 

The general equation of a circle

Not all circles are centred on the origin though! The following interactive allows you to explore the standard form equation of a circle. It shows how the equation changes as the coordinates of the centre $\left(h,k\right)$(h,k) and the radius $r$r change. To move the circle, drag the sliders for $h$h and $k$k, the centre coordinates of the circle, while to change the radius, just drag the $r$r slider. You'll notice that, regardless of the values, the equation will always be in the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2.

 

Equation of a circle in standard form

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

where $\left(h,k\right)$(h,k) is the coordinates of the centre of the circle

and $r$r is the radius of the circle

Equation of a circle in general form

The standard form of a circle can also be rearranged and written in general form. The general form of a circle is:

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

If we are given the equation of a circle in general form (or some variation), we will need to complete the square for $x$x and $y$y to find the centre and the radius and convert it into the standard form. This will also assist with sketching of the circle.

Worked example

Example 1

Given the expression $x^2+2x-6y+y^2-6=0$x2+2x6y+y26=0 which is thought to define a circle, 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. Thus:

$(x^2+2x+1)+(y^2-6y+9)$(x2+2x+1)+(y26y+9) $=$= $6+1+9$6+1+9
$(x+1)^2+(y-3)^2$(x+1)2+(y3)2 $=$= $16$16
$(x+1)^2+(y-3)^2$(x+1)2+(y3)2 $=$= $4^2$42

Hence this circle has centre $(-1,3)$(1,3) and radius $4$4.

Practice questions

Question 1

Consider the circle with centre ($0$0, $0$0) that passes through the point ($5$5, $-10$10).

  1. Find the exact radius of the circle.

  2. Determine the equation of the circle.

Question 2

The equation of a circle is given by $x^2+\left(y-3\right)^2=9$x2+(y3)2=9.

  1. State the coordinates of the centre of this circle.

  2. What is the radius of the circle?

  3. Find the $y$y values of the $y$y-intercepts. State your solutions on the same line, separated by a comma.

  4. Find the $x$x value of the $x$x-intercept.

  5. Plot the graph for the given circle.

    Loading Graph...

Question 3

Consider the equation of a circle given by $x^2+y^2-2x-10y-24=0$x2+y22x10y24=0.

  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? Express your answer in simplest surd form.

  4. Find the $y$y coordinates of the $y$y-intercepts. State your answers on the same line, separated by a comma.

  5. Find the $x$x coordinates of the $x$x-intercepts. State your answers on the same line, separated by a comma.

  6. Graph this circle on the axes below:

    Loading Graph...

 

The equation of a semicircle with centre at the origin

A circle is not a function, but it can be split into two semicircles, each of which are functions. The simplest case is for the circle $x^2+y^2=r^2$x2+y2=r2 with centre at the origin and radius $r$r.

For example, let's look at the circle:

$x^2+y^2$x2+y2 $=$= $9$9

Rearranging to make $y$y the subject, we get:

$y$y $=$= $\pm\sqrt{9-x^2}$±9x2
 

The diagram below shows how these two equations, $y=\sqrt{9-x^2}$y=9x2 and $y=-\sqrt{9-x^2}$y=9x2, define the top and bottom halves of a circle respectively.

 

 

Semicircles with centre at the origin
Equation Centre Radius Domain Range
$y=\sqrt{r^2-x^2}$y=r2x2 $\left(0,0\right)$(0,0) $r$r $[-r,r]$[r,r] $[0,r]$[0,r]
$y=-\sqrt{r^2-x^2}$y=r2x2 $\left(0,0\right)$(0,0) $r$r $[-r,r]$[r,r] $[-r,0]$[r,0]

 

The general equation of a semicircle

More generally, we can rearrange the equation of any circle in standard form to find two semicircles:

$\left(x-h\right)^2+\left(y-k\right)^2$(xh)2+(yk)2 $=$= $r^2$r2
$\left(y-k\right)^2$(yk)2 $=$= $r^2-\left(x-h\right)^2$r2(xh)2
$y-k$yk $=$= $\pm\sqrt{r^2-\left(x-h\right)^2}$±r2(xh)2
$y$y $=$= $k\pm\sqrt{r^2-\left(x-h\right)^2}$k±r2(xh)2
 

For example, a circle whose centre is located at $\left(2,5\right)$(2,5) and has radius $r=3$r=3 has the equation given by:

$y$y $=$= $5\pm\sqrt{9-\left(x-2\right)^2}$5±9(x2)2

It can be split into the two functions as follows:

$f\left(x\right)$f(x) $=$= $5+\sqrt{9-\left(x-2\right)^2}$5+9(x2)2

And:

$g\left(x\right)$g(x) $=$= $5-\sqrt{9-\left(x-2\right)^2}$59(x2)2
 

Each of these functions are semicircles. Both functions will have the same domain which is that of the original circle, given by $-1\le x\le5$1x5. The range of $f$f becomes $5\le y\le8$5y8 and the range of $g$g becomes $2\le y\le5$2y5 as shown here.


We can summarise this as follows:

General equation of semicircles
Equation Centre Radius Domain Range
$y=\sqrt{r^2-\left(x-h\right)^2}+k$y=r2(xh)2+k $\left(h,k\right)$(h,k) $r$r $\left[h-r,h+r\right]$[hr,h+r] $\left[k,k+r\right]$[k,k+r]
$y=-\sqrt{r^2-\left(x-h\right)^2}+k$y=r2(xh)2+k $\left(h,k\right)$(h,k) $r$r $\left[h-r,h+r\right]$[hr,h+r] $\left[k-r,k\right]$[kr,k]

 

Practice questions

Question 4

The following graph shows a semicircle.

Loading Graph...

  1. State the centre of the semicircle.

  2. State the radius of the semicircle.

    Radius = $\editable{}$ units

  3. Find the equation of this semicircle.

Question 5

A certain curve is described by the function $y=-\sqrt{16-x^2}$y=16x2.

  1. What kind of curve is it?

    Ellipse

    A

    Hyperbola

    B

    Parabola

    C

    Semicircle

    D
  2. State the centre of the semicircle.

  3. State the radius of the semicircle.

  4. Plot the graph for $y=-\sqrt{16-x^2}$y=16x2.

    Loading Graph...

Question 6

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

MA11-1

uses algebraic and graphical techniques to solve, and where appropriate, compare alternative solutions to problems

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