Circles
Understand the equation of a circle for Queensland Year 11 Mathematical Methods (QCAA). A circle is the set of points a fixed distance, the radius, from a centre.
You will learn to read the centre and radius from the standard equation, complete the square to get there from an expanded form, write a circle's equation from given information, and find its intercepts — another key relation in your gallery of graphs.
Every question with a fully worked solution.
- Circles - Video - Graphs of circles Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a circle has the equation \((x-h)^2+(y-k)^2=r^2\), with centre \((h,\,k)\) and radius \(r\). This page shows how to read the centre and radius, complete the square to find them, write a circle's equation, and find intercepts.
A circle is the set of all points a fixed distance \(r\), the radius, from a fixed point \((h,\,k)\), the centre. Its equation is \((x-h)^2+(y-k)^2=r^2\). When the centre is the origin this reduces to \(x^2+y^2=r^2\).
If the equation is expanded, such as \(x^2+y^2+dx+ey+f=0\), you recover the centre and radius by completing the square in \(x\) and in \(y\).
A circle fails the vertical-line test (most vertical lines cut it twice), so it is a relation, not a function.
The standard form of a circle:
The radius from the centre \((h,\,k)\) to a point \((x_1,\,y_1)\) on the circle:
How to find a circle's centre and radius
- Standard form: match to \((x-h)^2+(y-k)^2=r^2\) and read the centre \((h,\,k)\) and \(r=\sqrt{r^2}\).
- Expanded form: group the \(x\)-terms and \(y\)-terms and complete the square on each, moving the constant to the right.
- From centre and a point: use \(r^2=(x_1-h)^2+(y_1-k)^2\), then write the equation.
Match to standard form:
| \((x-0)^2+(y-0)^2\) | \(=\) | \(5^2\) |
| \(\text{centre}\) | \(=\) | \((0,\,0)\) |
| \(r\) | \(=\) | \(5\) |
\(x\)-intercepts — put \(y=0\):
| \(x^2\) | \(=\) | \(25\) |
| \(x\) | \(=\) | \(\pm 5\) |
\(y\)-intercepts — put \(x=0\):
| \(y^2\) | \(=\) | \(25\) |
| \(y\) | \(=\) | \(\pm 5\) |
Centre \((0,\,0)\), radius \(5\); intercepts \((\pm 5,\,0)\) and \((0,\,\pm 5)\).
Centre and radius:
| \(\text{centre}\) | \(=\) | \((3,\,-2)\) |
| \(r\) | \(=\) | \(\sqrt{25}=5\) |
Test \((7,\,1)\) — substitute into the left side:
| \((7-3)^2+(1+2)^2\) | \(=\) | \(4^2+3^2\) |
| \(=\) | \(16+9\) | |
| \(=\) | \(25\) |
The left side equals \(25=r^2\), so the point satisfies the equation.
Centre \((3,\,-2)\), radius \(5\); yes, \((7,\,1)\) lies on the circle.
Group \(x\)- and \(y\)-terms:
| \((x^2-6x)+(y^2+4y)\) | \(=\) | \(3\) |
Complete the square in each bracket:
| \((x-3)^2-9+(y+2)^2-4\) | \(=\) | \(3\) |
| \((x-3)^2+(y+2)^2\) | \(=\) | \(3+9+4\) |
| \((x-3)^2+(y+2)^2\) | \(=\) | \(16\) |
Read off centre and radius:
| \(\text{centre}\) | \(=\) | \((3,\,-2)\) |
| \(r\) | \(=\) | \(\sqrt{16}=4\) |
Centre \((3,\,-2)\), radius \(4\).
Radius — distance from centre to the point:
| \(r^2\) | \(=\) | \((2-(-1))^2+(6-2)^2\) |
| \(=\) | \(3^2+4^2\) | |
| \(=\) | \(9+16\) | |
| \(=\) | \(25\) |
Equation — substitute \(h=-1,\ k=2,\ r^2=25\):
| \((x-(-1))^2+(y-2)^2\) | \(=\) | \(25\) |
| \((x+1)^2+(y-2)^2\) | \(=\) | \(25\) |
Equation: \((x+1)^2+(y-2)^2=25\).
Common pitfalls
Frequently asked questions
What is the equation of a circle?
A circle with centre \((h,\,k)\) and radius \(r\) has equation \((x-h)^2+(y-k)^2=r^2\).
How do you find the centre and radius of a circle?
Match the equation to \((x-h)^2+(y-k)^2=r^2\): the centre is \((h,\,k)\) and the radius is \(\sqrt{r^2}\). If the equation is expanded, complete the square first.
How do you complete the square to find a circle's centre?
Group the \(x\)- and \(y\)-terms, add and subtract \(\left(\tfrac{d}{2}\right)^2\) for each, then rewrite as squared brackets and move the constant to the right.
Is a circle a function?
No. A vertical line through the interior cuts the circle twice, so it fails the vertical-line test and is a relation, not a function.
How do you write the equation of a circle from its centre and a point on it?
Find \(r^2=(x_1-h)^2+(y_1-k)^2\) using the given point, then substitute \(h,\,k,\,r^2\) into \((x-h)^2+(y-k)^2=r^2\).