Graphs of Circles
Theory
A circle with centre \((h,\ k)\) and radius \(r\) has equation \((x - h)^2 + (y - k)^2 = r^2\); with centre the origin it is \(x^2 + y^2 = r^2\). This Year 10A maths page derives the equation from the distance formula, finds the centre and radius by completing the square, and solves problems about intercepts, points inside or outside, shortest distances, and lines that cut or touch a circle.
A circle is the set of all points that are the same distance, the radius \(r\), from a fixed point, the centre.
If \(P(x,\ y)\) is on the circle with centre \(O(0,\ 0)\), the distance formula gives \(\sqrt{x^2 + y^2} = r\). Squaring both sides gives \(x^2 + y^2 = r^2\). With centre \((h,\ k)\) the same steps give \((x - h)^2 + (y - k)^2 = r^2\).
The general form \(x^2 + y^2 + ax + by + c = 0\) is the same circle with the brackets expanded. Completing the square on \(x\) and on \(y\) turns it back into centre-radius form.
A circle is not a function: a vertical line through its centre meets it twice. The top half is \(y = \sqrt{r^2 - x^2}\) and the bottom half \(y = -\sqrt{r^2 - x^2}\).
Centre \((h,\ k)\), radius \(r\):
Distance formula and midpoint, used to find a radius or a centre:
How to find the centre and radius
- Tidy the equation: all \(x\) and \(y\) terms on the left, the constant on the right. If \(x^2\) and \(y^2\) have a coefficient such as \(2\), divide every term by it first.
- Complete each square: add \(\left(\dfrac{\text{coefficient}}{2}\right)^2\) for \(x\) and for \(y\) to both sides.
- Read the centre (signs reversed from the brackets) and \(r = \sqrt{\text{right-hand side}}\).
Problem types
- Intercepts: put \(y = 0\) for the \(x\)-intercepts and \(x = 0\) for the \(y\)-intercepts, then solve; exact answers may be surds.
- Equation from given information: centre and a point (radius by the distance formula); the ends of a diameter (centre is the midpoint, radius is half the length). Three points that make a right angle at one of them: the other two are the ends of a diameter (angle in a semicircle).
- Line and circle: substitute the line into the circle, solve the quadratic, then find each \(y\). One repeated solution means the line is a tangent.
- Other curves and circles: for \(y = x^2\), replace \(x^2\) by \(y\) and reject any \(y \lt 0\). For \(y = \dfrac{k}{x}\), substitute and multiply through by \(x^2\) to get a quadratic in \(x^2\).
- Tangent at a point: find the gradient of the radius, take the negative reciprocal, and use point-gradient form (Example 4).
- Unknown constant: complete the squares with the constant in place; the right-hand side is \(r^2\), so it must be positive.
- Practical problems: "every point within \(5\) m of \(S\)" is a circle of radius \(5\) about \(S\); a path or fence is a line to intersect with it.
Group the \(x\) terms and the \(y\) terms, move the constant across, and complete each square, adding the same numbers to both sides:
\((x^2 + 6x + 9) + (y^2 - 8y + 16)\) \(= -9 + 9 + 16\)
| \((x + 3)^2 + (y - 4)^2\) | \(=\) | \(16\) |
Centre \(\mathbf{(-3,\ 4)}\), radius \(\sqrt{16} = \mathbf{4}\).
\(x\)-intercepts, \(y = 0\): \((x + 3)^2 + 16 = 16\), so \((x + 3)^2 = 0\) and \(x = -3\) only. The circle touches the \(x\)-axis (the centre is \(4\) above it, exactly one radius).
| \(y\text{-intercepts: } 9 + (y - 4)^2\) | \(=\) | \(16\) |
| \((y - 4)^2\) | \(=\) | \(7\) |
| \(y\) | \(=\) | \(4 \pm \sqrt{7}\) |
(i) Find its equation.
(ii) Do the points \(A(7,\ 7)\) and \(B(-9,\ 10)\) lie inside, on or outside it?
(i) The radius is the distance from the centre to the point:
| \(r\) | \(=\) | \(\sqrt{(5 + 1)^2 + (11 - 3)^2}\) |
| \(\) | \(=\) | \(\sqrt{36 + 64} = 10\) |
| \((x + 1)^2 + (y - 3)^2\) | \(=\) | \(100\) |
(ii) Substitute each point into the left-hand side and compare with \(r^2 = 100\):
| \(A:\ (7 + 1)^2 + (7 - 3)^2\) | \(=\) | \(64 + 16\) |
| \(\) | \(=\) | \(80 \lt 100\) |
| \(B:\ (-9 + 1)^2 + (10 - 3)^2\) | \(=\) | \(64 + 49\) |
| \(\) | \(=\) | \(113 \gt 100\) |
\(A\) is inside the circle (closer than \(10\) to the centre) and \(B\) is outside it.
Substitute the line into the circle, so there is only one unknown:
| \(x^2 + (2x - 5)^2\) | \(=\) | \(25\) |
| \(x^2 + 4x^2 - 20x + 25\) | \(=\) | \(25\) |
| \(5x^2 - 20x\) | \(=\) | \(0\) |
| \(5x(x - 4)\) | \(=\) | \(0\) |
\(x = 0\) gives \(y = -5\); \(x = 4\) gives \(y = 3\). The line meets the circle at \(\mathbf{(0,\ -5)}\) and \(\mathbf{(4,\ 3)}\).
Check \((4,\ 3)\): \(16 + 9 = 25\).
\(5^2 + (-12)^2 = 25 + 144 = 169\), so \(P\) is on the circle. The tangent is perpendicular to the radius \(OP\):
| \(m_{OP}\) | \(=\) | \(\dfrac{-12 - 0}{5 - 0} = -\dfrac{12}{5}\) |
| \(m_{\text{tangent}}\) | \(=\) | \(\dfrac{5}{12}\quad (m_1 m_2 = -1)\) |
Point-gradient form through \(P(5,\ -12)\):
| \(y + 12\) | \(=\) | \(\dfrac{5}{12}(x - 5)\) |
| \(12y + 144\) | \(=\) | \(5x - 25\) |
| \(5x - 12y\) | \(=\) | \(\boldsymbol{169}\) |
Check: \(5(5) - 12(-12) = 25 + 144 = 169\).
Common pitfalls
Frequently asked questions
What is the equation of a circle?
A circle with centre \((h,\ k)\) and radius \(r\) has equation \((x - h)^2 + (y - k)^2 = r^2\). With centre the origin it is \(x^2 + y^2 = r^2\).
How do you find the centre and radius from the general form?
Group the x terms and the y terms, complete the square for each, and add the same numbers to the right-hand side. Then read the centre from the brackets and take the square root of the right-hand side for the radius.
How do you tell if a point is inside a circle?
Substitute the point into (x - h) squared plus (y - k) squared. If the result is less than r squared the point is inside, if it equals r squared it is on the circle, and if it is greater it is 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 substitute back to find the other coordinate. Two solutions mean two points, one means the line is a tangent, none means they do not meet.
Is a circle a function?
No. A vertical line through the inside of a circle meets it twice, so it fails the vertical line test. The top and bottom halves separately are functions.