Graphs Of Circles
Every question with a fully worked solution.
- Graphs Of Circles - Video - Graph x^2 + y^2 = 4 Watch
Theory
The equation of a circle comes straight from Pythagoras' theorem. This Year 10 maths page shows how to read the centre and radius from \(x^2 + y^2 = r^2\) and \((x - a)^2 + (y - b)^2 = r^2\), how to complete the square to find the centre of an expanded equation, how to test whether a point is inside, on or outside a circle, and how to find the intercepts and sketch the circle.
A circle is the set of all points that are the same distance, the radius \(r\), from a fixed point, the centre.
A circle with centre at the origin \((0, 0)\) and radius \(r\) has equation \(x^2 + y^2 = r^2\). It crosses the axes at \((\pm r, 0)\) and \((0, \pm r)\).
A circle with centre \((a, b)\) and radius \(r\) has equation \((x - a)^2 + (y - b)^2 = r^2\).
A point lies on a circle if its coordinates make the equation true. If the left side is smaller than \(r^2\), the point is inside; if it is larger, the point is outside.
An equation is a circle when \(x^2\) and \(y^2\) are both present, added, with the same coefficient. \(x^2 - y^2 = 9\) and \(y = x^2 + 9\) are not circles.
The two forms of the equation of a circle:
How to find the centre and radius
- If the equation is \((x - a)^2 + (y - b)^2 = r^2\), read the centre \((a, b)\): change the sign of each number in the brackets. The radius is the square root of the right side.
- If it is expanded, group the \(x\) terms and the \(y\) terms. Complete the square for each: add (half the coefficient)\(^2\) to both sides.
- Write it in the form \((x - a)^2 + (y - b)^2 = r^2\) and read off the centre and radius.
Intercepts, points and sketching
- \(x\)-intercepts: put \(y = 0\) and solve for \(x\); \(y\)-intercepts: put \(x = 0\). Take \(\pm\) square roots and leave any surd exact.
- Centre and a point: the radius is the distance from the centre to the point (Pythagoras or the distance formula).
- A horizontal or vertical line such as \(y = 6\): substitute it and solve for the other coordinate.
- Sketch: plot the centre, mark points one radius left, right, above and below it, and draw the circle through them. Label the intercepts.
(i) Find the radius of \(x^2 + y^2 = 81\) and where it crosses the axes.
(ii) Does the point \((-2, 5)\) lie on the circle \(x^2 + y^2 = 30\)?
(i) Compare with \(x^2 + y^2 = r^2\).
| \(r^2\) | \(=\) | \(81\) |
| \(r\) | \(=\) | \(\mathbf{9}\) |
| \(\text{intercepts}\) | \(\) | \(\boldsymbol{(\pm 9, 0)} \text{ and } \boldsymbol{(0, \pm 9)}\) |
(ii) Substitute the point. It lies on the circle only if the left side equals \(30\).
| \((-2)^2 + 5^2\) | \(=\) | \(4 + 25 = 29\) |
| \(29\) | \(\ne\) | \(30\) |
No. Since \(29 < 30\), the point is just inside the circle.
(i) State the centre and radius of \((x - 4)^2 + (y + 1)^2 = 36\).
(ii) Write the equation of the circle with centre \((-5, 2)\) and radius \(4\).
(i) Compare with \((x - a)^2 + (y - b)^2 = r^2\). \(y + 1 = y - (-1)\), so \(b = -1\).
| \(\text{centre}\) | \(\) | \(\boldsymbol{(4, -1)}\) |
| \(r\) | \(=\) | \(\sqrt{36} = \mathbf{6}\) |
(ii) Put \(a = -5\), \(b = 2\), \(r = 4\). Subtracting \(-5\) gives \(+5\).
| \((x - (-5))^2 + (y - 2)^2\) | \(=\) | \(4^2\) |
| \(\) | \(\) | \(\boldsymbol{(x + 5)^2}\) \(\boldsymbol{{} + (y - 2)^2 = 16}\) |
Complete the square for the \(x\) terms and for the \(y\) terms. Add the same numbers to the right side.
Half of \(-8\) is \(-4\), and \((-4)^2 = 16\). Half of \(4\) is \(2\), and \(2^2 = 4\).
| \((x^2 - 8x + 16) + (y^2 + 4y + 4)\) | \(=\) | \(5 + 16 + 4\) |
| \((x - 4)^2 + (y + 2)^2\) | \(=\) | \(25\) |
| \(\text{centre}\) | \(\) | \(\boldsymbol{(4, -2)}\) |
| \(r\) | \(=\) | \(\sqrt{25} = \mathbf{5}\) |
(i) Find its equation.
(ii) Find its exact \(x\)- and \(y\)-intercepts, and sketch it.
(i) The radius is the distance from the centre to the point.
| \(r^2\) | \(=\) | \((3 - (-2))^2 + (2 - 1)^2\) |
| \(\) | \(=\) | \(25 + 1 = 26\) |
| \(r\) | \(=\) | \(\sqrt{26} \approx 5.1\) |
Equation: \(\boldsymbol{(x + 2)^2 + (y - 1)^2 = 26}\).
(ii) \(x\)-intercepts: put \(y = 0\).
| \((x + 2)^2 + 1\) | \(=\) | \(26\) |
| \((x + 2)^2\) | \(=\) | \(25\) |
| \(x + 2\) | \(=\) | \(\pm 5\) |
| \(x\) | \(=\) | \(\boldsymbol{3} \text{ or } \boldsymbol{-7}\) |
\(y\)-intercepts: put \(x = 0\).
| \(4 + (y - 1)^2\) | \(=\) | \(26\) |
| \((y - 1)^2\) | \(=\) | \(22\) |
| \(y\) | \(=\) | \(\boldsymbol{1 \pm \sqrt{22}} \approx 5.69 \text{ or } -3.69\) |
Common pitfalls
Frequently asked questions
What is the equation of a circle?
A circle with centre \((a, b)\) and radius \(r\) has equation \((x - a)^2 + (y - b)^2 = r^2\). With centre at the origin it is \(x^2 + y^2 = r^2\).
Why does the equation come from Pythagoras' theorem?
Any point on the circle is a distance \(r\) from the centre. The horizontal and vertical distances to the centre form a right-angled triangle with hypotenuse \(r\).
How do you check if a point is on a circle?
Substitute the coordinates into the equation. If both sides are equal the point is on the circle. Smaller means inside, larger means outside.
How do you find the centre of an expanded circle equation?
Complete the square for the x terms and the y terms, adding the same numbers to the right side, then read the centre and radius from the new form.
How many times can a line meet a circle?
Twice, once (when the line just touches the circle, a tangent) or not at all.