Analytic Geometry
The Circle and Its Equation
The circle as a locus, its equation from Pythagoras, and recovering centre and radius from an expanded form by completing the square.
What this page covers
- Derive the equation of a circle from the distance formula
- Write the equation given a centre and radius
- Recover centre and radius from an expanded equation
- Recognise when an equation describes no circle at all
The circle as a locus
A circle is the path of a point which moves so that its distance from a fixed point is constant. The fixed point is the centre, the constant distance the radius.
That wording is the source's own, and it is worth using because it makes the derivation immediate: the definition is a statement about distance, and there is a formula for distance.
Deriving the equation, centre at the origin
Let (x, y) be any point at distance r from the origin. The distance formula gives
The source draws the right triangle explicitly: legs x and y, hypotenuse r. The equation is the theorem.
Centre (-5, 1) gives (x + 5)2 + (y - 1)2 = r2. The h and k appear with reversed signs because the formula subtracts them. Reading the centre straight off the brackets without flipping is a persistent error.
Reading a circle off its equation
| Equation | Centre | Radius |
|---|---|---|
| x2 + y2 = 4 | (0, 0) | 2 |
| (x - 2)2 + (y - 3)2 = 4 | (2, 3) | 2 |
| (x + 5)2 + (y - 1)2 = 5 | (-5, 1) | √5 |
The first two have the same radius and differ only by a translation, which the source draws side by side. Replacing x by x - h shifts the graph h to the right; replacing y by y - k shifts it k upward.
The right side is r2, not r. In (x + 5)2 + (y - 1)2 = 5 the radius is √5, about 2.24, not 5. The source states this explicitly.
Recovering the centre from an expanded equation
An expanded circle equation hides its centre. Completing the square on x and y separately recovers it.
Worked example — the source's case
Find the centre and radius of x2 + 2x + y2 + 4y = 7.
- Group. (x2 + 2x) + (y2 + 4y) = 7.
- Complete each square. For x: half of 2 is 1, squared is 1. For y: half of 4 is 2, squared is 4.
- Add to both sides. 1 + 4 = 5 goes on the right as well.
Substitute the point directly above the centre, (-1, -2 + 2√3): 0 + (2√3)2 = 12 ✓.
The general form of a circle is x2 + y2 + Dx + Ey + F = 0, and completing the square always reduces it to standard form. Two features identify it before any work is done: the x2 and y2 coefficients are equal, and there is no xy term.
Degenerate cases
Completing the square can leave a right-hand side that is zero or negative, and neither gives a circle.
| Result | Locus | Reason |
|---|---|---|
| (x - h)2 + (y - k)2 = r2, r > 0 | A circle of radius r | The general case |
| (x - h)2 + (y - k)2 = 0 | The single point (h, k) | A sum of squares is zero only if both are |
| (x - h)2 + (y - k)2 < 0 | Nothing at all | A sum of squares is never negative |
Checking the sign of the right side after completing the square is a one-second test that prevents a great deal of pointless algebra.
Symmetry and the vertical line test
The source notes that a circle centred at the origin has symmetry about the y-axis, the x-axis, the origin and the line y = x — all four at once, which is a strong hint at how symmetric the object is.
It also observes that a circle is not a function. For x2 + y2 = 9, any x strictly between -3 and 3 gives two y values. The source draws a vertical line cutting the circle twice, which is the vertical line test failing.
Splitting into y = √9 - x2 and y = -√9 - x2 gives two genuine functions, the upper and lower semicircles. That is the standard way of handling a relation that is not a function.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Reading (x + 5)2 as centre x = 5 | Centre x = -5 | The formula subtracts h |
| Taking the right side as the radius | It is r2 | = 5 means r = √5 |
| Completing the square without balancing | Add to both sides | The equation must stay true |
| Completing only the x terms | Both variables | The centre has two coordinates |
| Calling a negative right side a circle | No locus | A sum of squares is never negative |
| Treating a circle as a function | It is a relation | Vertical line test fails |
Frequently asked questions
Why does the equation of a circle look like Pythagoras?
Because it is. A point (x, y) lies on the circle exactly when its distance from the centre equals r, and that distance is computed by Pythagoras. Squaring both sides removes the radical.
How do I tell a circle from other second-degree equations?
The x2 and y2 coefficients must be equal and non-zero, and there must be no xy term. Unequal coefficients give an ellipse; opposite signs give a hyperbola.
What if completing the square gives a negative on the right?
There is no such circle. (x - h)2 + (y - k)2 = -4 has no real solutions, because a sum of squares is never negative. A right side of zero gives the single point (h, k).
Is a circle a function?
No. It fails the vertical line test: a vertical line through the interior meets it twice. The source uses x2 + y2 = 9 as its example of a relation that is not a function.
Source. Handwritten teaching notes, Week 4, pages 11-12.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.