Graphs of Circles
The circle centred at the origin with radius \(r\) has equation \(x^2 + y^2 = r^2\), which comes from the distance formula. This NSW Year 9 Advanced maths subtopic graphs circles and moves on to circles with any centre (NESA outcome MA5‑NLI‑P‑01).
You will write equations from a centre and radius, sketch circles, decide whether points lie inside, on or outside a circle, and find where lines meet a circle. You will use \((x - a)^2 + (y - b)^2 = r^2\), complete the square to find a centre and radius, find a circle from a diameter, map a phone tower's coverage, and prove that a line misses a circle. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
Theory
A circle is every point the same distance, the radius, from its centre. Pythagoras' theorem turns that into an equation: \(x^2 + y^2 = r^2\) for centre the origin, and \((x - a)^2 + (y - b)^2 = r^2\) for centre \((a, b)\). This Year 9 Advanced maths page is part of the path towards Advanced and Extension maths.
A circle is the set of all points that are a fixed distance, the radius \(r\), from a fixed point, the centre.
For centre \(O(0, 0)\), a point \(P(x, y)\) is on the circle when its distance from \(O\) is \(r\). Pythagoras' theorem gives the equation \(x^2 + y^2 = r^2\). So \(x^2 + y^2 = 49\) is a circle of radius \(7\), and \(x^2 + y^2 = 12\) has radius \(\sqrt{12} = 2\sqrt{3}\).
For centre \((a, b)\), the equation is \((x - a)^2 + (y - b)^2 = r^2\). Watch the signs: \((x + 2)^2\) means \(a = -2\).
Expanded, a circle looks like \(x^2 + y^2 + px + qy + c = 0\): equal coefficients of \(x^2\) and \(y^2\), and no \(xy\) term. Completing the square turns it back into centre–radius form.
A point is on the circle if it satisfies the equation, inside if its distance from the centre is less than \(r\), and outside if it is more.
Centre the origin, radius \(r\):
Centre \((a, b)\), radius \(r\):
Distance and midpoint, used for radii, diameters and chords:
How to work with a circle's equation
- Make the squares monic: if \(x^2\) and \(y^2\) share a coefficient, divide every term by it first.
- Read or find the centre and radius: compare with \((x - a)^2 + (y - b)^2 = r^2\). If the equation is expanded, complete the square in \(x\) and in \(y\), adding the same numbers to both sides.
- Radius from a point: use the distance from the centre to any point on the circle. A diameter's midpoint is the centre and half its length is the radius.
- Intercepts: put \(y = 0\) for \(x\)-intercepts, \(x = 0\) for \(y\)-intercepts. Keep both square roots, \(\pm\), and simplify surds.
- Sketch: mark the centre, then the four points one radius left, right, above and below it, and draw the circle through them.
Lines, points and unknowns
- Line meets circle: substitute the line into the circle, solve, then find each \(y\) from the line. A horizontal line \(y = k\) needs only one substitution.
- Chord length: the distance between the two meeting points.
- Does it cross an axis? Compare the distance from the centre to the axis with the radius.
- Unknown centre: substitute each known point into the equation; keep the value that works for all of them.
(i) Find its equation.
(ii) Find the points on the circle where \(x = 6\).
(iii) Is \((7, 7)\) inside, on or outside the circle?
(i) The radius is the distance from \(O\) to the point.
| \(r\) | \(=\) | \(\sqrt{8^2 + (-6)^2} = \sqrt{100} = 10\) |
| \(x^2 + y^2\) | \(=\) | \(\mathbf{100}\) |
(ii) Substitute \(x = 6\) and take both square roots.
| \(36 + y^2\) | \(=\) | \(100\) |
| \(y^2\) | \(=\) | \(64\) |
| \(y\) | \(=\) | \(\pm 8\) |
The points are \(\mathbf{(6, 8)}\) and \(\mathbf{(6, -8)}\).
(iii) Compare \(x^2 + y^2\) with \(r^2 = 100\): \(7^2 + 7^2 = 98 < 100\), so \((7, 7)\) is inside, just.
(i) Write the equation of the circle with centre \((-3, 2)\) and radius \(5\), and expand it into the form \(x^2 + y^2 + px + qy + c = 0\).
(ii) Find where the circle crosses the \(y\)-axis.
(i) Use \((x - a)^2 + (y - b)^2 = r^2\) with \(a = -3\), \(b = 2\). Subtracting \(-3\) makes \(x + 3\).
| \((x + 3)^2 + (y - 2)^2\) | \(=\) | \(25\) |
| \(x^2 + 6x + 9 + y^2 - 4y + 4\) | \(=\) | \(25\) |
| \(x^2 + y^2 + 6x - 4y - 12\) | \(=\) | \(\mathbf{0}\) |
(ii) On the \(y\)-axis, \(x = 0\). The bracket form is quicker:
| \(9 + (y - 2)^2\) | \(=\) | \(25\) |
| \((y - 2)^2\) | \(=\) | \(16\) |
| \(y - 2\) | \(=\) | \(\pm 4\) |
| \(y\) | \(=\) | \(\mathbf{6} \text{ or } \mathbf{-2}\) |
(i) Show that \(x^2 + y^2 - 8x + 2y + 8 = 0\) is a circle and find its centre and radius.
(ii) Find the exact \(x\)-intercepts.
(i) Group the \(x\) terms and the \(y\) terms. Add the square of half of each coefficient to both sides.
| \((x^2 - 8x + 16) + (y^2 + 2y + 1)\) | \(=\) | \(-8 + 16 + 1\) |
| \((x - 4)^2 + (y + 1)^2\) | \(=\) | \(9\) |
Centre \(\mathbf{(4, -1)}\), radius \(\sqrt{9} = \mathbf{3}\).
(ii) The centre is \(1\) unit below the \(x\)-axis and the radius is \(3\), so the circle does cross it. Put \(y = 0\):
| \((x - 4)^2 + 1\) | \(=\) | \(9\) |
| \((x - 4)^2\) | \(=\) | \(8\) |
| \(x - 4\) | \(=\) | \(\pm\sqrt{8} = \pm 2\sqrt{2}\) |
| \(x\) | \(=\) | \(\mathbf{4 \pm 2\sqrt{2}}\) |
Substitute the line into the circle, then solve the quadratic.
| \(x^2 + (x - 2)^2\) | \(=\) | \(10\) |
| \(2x^2 - 4x + 4\) | \(=\) | \(10\) |
| \(x^2 - 2x - 3\) | \(=\) | \(0\) |
| \((x - 3)(x + 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3 \text{ or } x = -1\) |
Find each \(y\) from the line: \(\mathbf{A(3, 1)}\) and \(\mathbf{B(-1, -3)}\).
| \(AB\) | \(=\) | \(\sqrt{(3 - (-1))^2 + (1 - (-3))^2}\) |
| \(\) | \(=\) | \(\sqrt{32} = \mathbf{4\sqrt{2}}\text{ units}\) |
Check \(A\) on the circle: \(9 + 1 = 10\) \(\checkmark\)
Common pitfalls
Frequently asked questions
What is the equation of a circle?
A circle with centre \((a, b)\) and radius \(r\) has equation \((x - a)^2 + (y - b)^2 = r^2\). If the centre is the origin this becomes \(x^2 + y^2 = r^2\).
How do you find the centre and radius of a circle?
Write the equation in the form \((x - a)^2 + (y - b)^2 = r^2\), completing the square if needed. The centre is \((a, b)\) and the radius is the square root of the number on the right.
Why is the equation of a circle x squared + y squared = r squared?
For any point \((x, y)\) on the circle, the horizontal distance \(x\), the vertical distance \(y\) and the radius \(r\) form a right-angled triangle. Pythagoras' theorem gives \(x^2 + y^2 = r^2\).
How do you tell if a point is inside a circle?
Find the point's distance from the centre, or for a circle centred at the origin work out \(x^2 + y^2\). If it is less than \(r^2\) the point is inside, equal means on the circle, and greater means outside.
How do you find where a line meets a circle?
Substitute the line's equation into the circle's equation to get a quadratic in one variable. Solve it, then use the line to find the other coordinate of each point. Two solutions means two meeting points, one means the line is a tangent, none means they do not meet.
How do you complete the square for a circle?
Group the \(x\) terms and the \(y\) terms. Add the square of half the \(x\) coefficient and the square of half the \(y\) coefficient to both sides. Then factor each group as a perfect square.