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.
$x^2+y^2=r^2$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 $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2.
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$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)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
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+2x−6y+y2−6=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)+(y2−6y+9) | $=$= | $6+1+9$6+1+9 |
| $(x+1)^2+(y-3)^2$(x+1)2+(y−3)2 | $=$= | $16$16 |
| $(x+1)^2+(y-3)^2$(x+1)2+(y−3)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).
Find the exact radius of the circle.
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+(y−3)2=9.
State the coordinates of the centre of this circle.
What is the radius of the circle?
Find the $y$y values of the $y$y-intercepts. State your solutions on the same line, separated by a comma.
Find the $x$x value of the $x$x-intercept.
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+y2−2x−10y−24=0.
Rewrite the equation of the circle in the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2.
What are the coordinates of the centre of this circle?
What is the radius of this circle? Express your answer in simplest surd form.
Find the $y$y coordinates of the $y$y-intercepts. State your answers on the same line, separated by a comma.
Find the $x$x coordinates of the $x$x-intercepts. State your answers on the same line, separated by a comma.
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 | $=$= | $\pm\sqrt{9-x^2}$±√9−x2 |
The diagram below shows how these two equations, $y=\sqrt{9-x^2}$y=√9−x2 and $y=-\sqrt{9-x^2}$y=−√9−x2, define the top and bottom halves of a circle respectively.
| Equation | Centre | Radius | Domain | Range |
|---|---|---|---|---|
| $y=\sqrt{r^2-x^2}$y=√r2−x2 | $\left(0,0\right)$(0,0) | $r$r | $[-r,r]$[−r,r] | $[0,r]$[0,r] |
| $y=-\sqrt{r^2-x^2}$y=−√r2−x2 | $\left(0,0\right)$(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:
| $\left(x-h\right)^2+\left(y-k\right)^2$(x−h)2+(y−k)2 | $=$= | $r^2$r2 |
| $\left(y-k\right)^2$(y−k)2 | $=$= | $r^2-\left(x-h\right)^2$r2−(x−h)2 |
| $y-k$y−k | $=$= | $\pm\sqrt{r^2-\left(x-h\right)^2}$±√r2−(x−h)2 |
| $y$y | $=$= | $k\pm\sqrt{r^2-\left(x-h\right)^2}$k±√r2−(x−h)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−(x−2)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−(x−2)2 |
And:
| $g\left(x\right)$g(x) | $=$= | $5-\sqrt{9-\left(x-2\right)^2}$5−√9−(x−2)2 |
Each of these functions are semicircles. Both functions will have the same domain which is that of the original circle, given by $-1\le x\le5$−1≤x≤5. The range of $f$f becomes $5\le y\le8$5≤y≤8 and the range of $g$g becomes $2\le y\le5$2≤y≤5 as shown here.
We can summarise this as follows:
| Equation | Centre | Radius | Domain | Range |
|---|---|---|---|---|
| $y=\sqrt{r^2-\left(x-h\right)^2}+k$y=√r2−(x−h)2+k | $\left(h,k\right)$(h,k) | $r$r | $\left[h-r,h+r\right]$[h−r,h+r] | $\left[k,k+r\right]$[k,k+r] |
| $y=-\sqrt{r^2-\left(x-h\right)^2}+k$y=−√r2−(x−h)2+k | $\left(h,k\right)$(h,k) | $r$r | $\left[h-r,h+r\right]$[h−r,h+r] | $\left[k-r,k\right]$[k−r,k] |
Practice questions
Question 4
The following graph shows a semicircle.
State the centre of the semicircle.
State the radius of the semicircle.
Radius = $\editable{}$ units
Find the equation of this semicircle.
Question 5
A certain curve is described by the function $y=-\sqrt{16-x^2}$y=−√16−x2.
What kind of curve is it?
Ellipse
AHyperbola
BParabola
CSemicircle
DState the centre of the semicircle.
State the radius of the semicircle.
Plot the graph for $y=-\sqrt{16-x^2}$y=−√16−x2.
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].
Plot the semicircle.
Loading Graph...State the equation for the semicircle in the form $y=\pm\sqrt{r^2-\left(x-h\right)^2}+k$y=±√r2−(x−h)2+k.