Circle Graph: x² + y² = r²
A circle is the set of points at a fixed distance from a centre. It is the first equation most students meet that cannot be written as y = f(x), which makes it a useful test of what a graphing tool can actually do.
Open this graph in the calculator
What you are looking at
The teal curve is x² + y² = 25, a circle of radius 5 centred at the origin. The 25 is r², so the radius is √25 = 5.
The rose curve is (x − 3)² + (y + 2)² = 4, centred at (3, −2) with radius 2. Note the signs: the centre coordinates are the opposite of what appears inside the brackets, which is the same rule as vertex form for a parabola.
The standard form
Any circle can be written as (x − h)² + (y − k)² = r², where (h, k) is the centre and r is the radius. Reading a circle off its equation is then immediate.
| Equation | Centre | Radius |
|---|---|---|
| x² + y² = 25 | (0, 0) | 5 |
| (x − 3)² + (y + 2)² = 4 | (3, −2) | 2 |
| (x + 1)² + y² = 9 | (−1, 0) | 3 |
Why a circle is not a function
A function returns exactly one output per input. A circle fails that: at x = 0 on the first circle, y could be 5 or −5. It fails the vertical line test, because a vertical line through the middle crosses it twice.
This is why you cannot type a circle as y = something. Solving for y gives y = ±√(25 − x²), which is two separate functions, an upper and a lower semicircle. You can graph those separately if you want to, and it is a useful exercise: each half is a function, and together they make the circle.
How the calculator draws it
Because a circle is not a function of x, the usual approach of stepping along the x-axis and evaluating cannot work. Instead the calculator rearranges the equation to the form f(x, y) = 0, evaluates that across a grid, and traces the contour where the result changes sign. The technique is called marching squares.
This is also what lets it draw any implicit equation you type, however unusual. Try x² − y² = 1 for a hyperbola, or something less orthodox like sin(x) = cos(y), which produces a lattice of curves.
Questions
How do I find the centre and radius from an expanded equation?
Complete the square in x and in y separately. x² + y² − 6x + 4y − 3 = 0 becomes (x − 3)² + (y + 2)² = 16, so the centre is (3, −2) and the radius is 4.
What happens if r² is negative?
There are no real solutions, so nothing is drawn. x² + y² = −4 has no points, because a sum of two squares cannot be negative.
What if r² is zero?
The graph is a single point at the centre. x² + y² = 0 is satisfied only at the origin. It is sometimes called a degenerate circle.