Equation of a circle

A point (x, y) is on the circle when its distance from the centre (h, k) is r. The distance formula says that distance is √((x − h)² + (y − k)²). Setting it equal to r and squaring both sides gives the equation above. The circle centred at the origin is the special case x² + y² = r².

As with vertex form, the signs inside the brackets are flipped: (x + 2)² + (y − 3)² = 16 has centre (−2, 3) and radius 4, the square root of 16, not 16.

Expanded, a circle’s equation looks like x² + y² + Dx + Ey + F = 0. To find the centre and radius, complete the square in x and in y. Take x² + y² − 6x + 4y − 12 = 0:

1. Group the x terms and the y terms: (x² − 6x) + (y² + 4y) = 12.
2. Complete each square by adding (half the coefficient)² to both sides: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4.
3. Factor: (x − 3)² + (y + 2)² = 25.

The centre is (3, −2) and the radius is 5. In general the centre is (−D/2, −E/2) and r² = D²/4 + E²/4 − F. If that last number is zero the “circle” is a single point, and if it is negative there are no points at all.

1. Plot the centre (h, k).
2. From the centre, count r units right, left, up and down and mark those four points.
3. Join them with a round curve, or use a compass set to radius r.

Solving x² + y² = 9 for y gives y = ±√(9 − x²): two values for most x, the top and bottom semicircles. A function can only have one y for each x, so a whole circle fails the vertical line test. Many calculators make you type the two halves separately. Here you can type the equation exactly as it is, x^2 + y^2 = 9, and it is drawn as an implicit curve.

• Parametric form: (h + r cos t, k + r sin t) for 0 ≤ t ≤ 2π traces the circle once.
• Polar form: r = 2a cos θ is a circle of radius a through the origin, centred at (a, 0).
• Inequalities: (x − h)² + (y − k)² < r² is the disc inside the circle.
• Ellipses: dividing by different amounts, x²/a² + y²/b² = 1, stretches the circle into an ellipse.