Equation of a Circle: From Completing the Square to Three-Point Circles

One circle, two costumes

Every circle in the coordinate plane can be written two ways. The standard form, (x − h)² + (y − k)² = r², wears its geometry on its sleeve: the centre is (h, k), the radius is r, and you can sketch the circle from the equation in seconds. The general form, x² + y² + Dx + Ey + F = 0, is the same equation after someone has multiplied everything out — the centre and radius are still in there, but hidden inside the coefficients.

Exams, textbooks and software are inconsistent about which form they hand you, so the real skill is moving between them without dropping a sign. The circle equation calculator does the conversion instantly from a centre and radius, and its page covers the definitions and the coefficient formulas. This guide covers everything around that: completing the square properly, building the equation when you are given a diameter or three points instead of a centre, graphing, and the places the algebra likes to bite.

Completing the square, without losing the plot

Going from standard form to general form is just expansion — multiply out the brackets, collect terms, move r² across. Going the other way is where marks are lost, so here it is in full. Take:

x² + y² + 6x − 2y − 15 = 0

Step 1 — group by variable, constants right. Put the x-terms together, the y-terms together, and move the loose constant to the other side: (x² + 6x) + (y² − 2y) = 15.

Step 2 — complete each square. Take half of each linear coefficient and square it. Half of 6 is 3, squared is 9. Half of −2 is −1, squared is 1. Add both numbers to both sides — this is the step people skip on the right-hand side: (x² + 6x + 9) + (y² − 2y + 1) = 15 + 9 + 1.

Step 3 — factor and read off. Each grouped trinomial is now a perfect square: (x + 3)² + (y − 1)² = 25. The centre is (−3, 1) — note the sign flips, because the form subtracts the coordinates — and the radius is √25 = 5, not 25.

There is also a shortcut that skips the algebra entirely: the centre is always (−D/2, −E/2) and the radius is √(D²/4 + E²/4 − F). With D = 6, E = −2, F = −15 that gives (−3, 1) and √(9 + 1 + 15) = 5, matching the long way. The shortcut is faster; completing the square is what you will be asked to show. Use the calculator in reverse to check either: enter the centre and radius you found, and confirm the general form it prints is the equation you started with.

Worked example: the circle hiding in a diameter

A very common exam setup gives you no centre and no radius — just the two endpoints of a diameter. Say the diameter runs from (−1, 4) to (5, −4).

The centre is the midpoint of the diameter, because the centre sits exactly halfway across any diameter: h = (−1 + 5)/2 = 2 and k = (4 + (−4))/2 = 0, so the centre is (2, 0).

The radius is half the diameter’s length. The distance between the endpoints is √((5 − (−1))² + (−4 − 4)²) = √(36 + 64) = √100 = 10, so r = 5.

Feeding centre (2, 0) and radius 5 into the circle equation calculator gives the standard form

(x − 2)² + y² = 25

and the general form x² + y² − 4x − 21 = 0. Two details worth noticing. Because k = 0, the y-bracket collapses to a bare y² and the general form has no y-term at all — E = −2k = 0, and the calculator simply drops it rather than printing “+ 0y”. And you can verify the whole thing by substituting an original endpoint: (−1 − 2)² + 4² = 9 + 16 = 25 ✓. If an endpoint of the diameter does not sit on your circle, something went wrong at the midpoint step.

Building the equation from other clues

Centre-and-radius is the friendly case. Real problems dress the same information up in different clothes.

Tangent to an axis

“Centre (4, 3), tangent to the x-axis” is a radius in disguise: tangent means the circle just touches the axis, so the radius equals the centre’s distance from that axis — here r = |k| = 3, giving (x − 4)² + (y − 3)² = 9. Tangent to the y-axis would instead mean r = |h| = 4. No distance formula needed; the coordinate you already haveis the answer.

Three points on the circle

Three non-collinear points determine exactly one circle — the circumcircle of the triangle they form. The cleanest route is the general form: each point substituted into x² + y² + Dx + Ey + F = 0 gives one linear equation in D, E and F. For the points (0, 0), (6, 0) and (0, 8):

• (0, 0): 0 + 0 + 0 + 0 + F = 0, so F = 0.
• (6, 0): 36 + 6D + F = 0, so D = −6.
• (0, 8): 64 + 8E + F = 0, so E = −8.

The circle is x² + y² − 6x − 8y = 0, which completes the square to (x − 3)² + (y − 4)² = 25: centre (3, 4), radius 5. There is a lovely check hiding here — the triangle with those three vertices is right-angled at the origin (its legs run along the axes, and 6-8-10 is twice the classic 3-4-5 triangle), and Thales’ theorem says a right angle inscribed in a circle always stands on a diameter. Sure enough, the hypotenuse from (6, 0) to (0, 8) has length 10 = 2r and its midpoint (3, 4) is the centre.

Graphing: why a circle is not a function

Type a circle equation into a function grapher and it will refuse, because a circle fails the vertical line test — most x-values correspond to two y-values. To plot one you solve for y and get two semicircles. For (x − 2)² + y² = 25:

y = ±√(25 − (x − 2)²)

The + branch is the top half, the − branch the bottom, and together they close the loop. This is also why graphing calculators draw circles with a small gap at the sides: near x = −3 and x = 7 the semicircle functions become vertical and the plotter runs out of sample points.

Intercepts come from the same move. Set y = 0 for x-intercepts: (x − 2)² = 25 gives x = 7 and x = −3. Set x = 0 for y-intercepts: y² = 25 − 4 = 21, so y = ±√21 ≈ ±4.58. A circle can have two, one or zero intercepts on each axis depending on how far the centre sits from it — compare the radius with |k| or |h| to know in advance, exactly as in the tangency shortcut above.

Spotting a circle among the conics

The general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 covers ellipses, parabolas and hyperbolas too. A circle is the special case with no xy-term and equal x² and y² coefficients (B = 0, A = C). If the squared coefficients match but are not 1, divide the whole equation through before doing anything else. For example, 4x² + 4y² − 8x + 16y − 29 = 0 becomes x² + y² − 2x + 4y − 7.25 = 0 after dividing by 4, which the coefficient shortcut turns into centre (1, −2) and radius √(1 + 4 + 7.25) = √12.25 = 3.5. Skip the division and the −D/2 formulas silently give the wrong centre.

One caveat the calculator page’s FAQ covers in more depth: passing the B = 0, A = C test does not guarantee a real circle. The expression D²/4 + E²/4 − F must come out positive; zero gives a single point and a negative value gives no graph at all. Always compute the radius before declaring victory.

Common mistakes

Adding to one side only. Completing the square means adding (D/2)² and (E/2)² to both sides. Add them on the left and forget the right, and every subsequent number — radius included — is wrong. In the walkthrough above, the 15 became 25 precisely because 9 and 1 crossed over too.

Reporting r² as the radius. The standard form ends in r², so (x + 3)² + (y − 1)² = 25 describes a circle of radius 5, not 25. It is the single most common slip on this topic, and it makes the circle 25 times too big in area.

Sign errors when halving. The centre is (−D/2, −E/2) — with the minus built in. From x² + y² + 6x − 2y − 15 = 0, the centre is (−3, 1): the +6x gives a negative h and the −2y a positive k. If your recovered centre has the same signs as the coefficients, you have almost certainly dropped the minus.

Forgetting to normalise first. Every formula on this page assumes the x² and y² coefficients are exactly 1. Given 3x² + 3y² + …, divide by 3 before halving anything — including the constant term, which is the one people miss.

When the equation is the beginning, not the end

The equation is usually a means to something else: once you hold the centre and radius, questions about circumference, area and arc length are one step away, and questions about whether a line clips the circle reduce to comparing the centre’s distance from the line with r. The coordinate techniques here — distance, midpoint, and the Pythagorean identity underneath them — are the same three tools throughout. For the definitions, the derivation and a second worked example, the circle equation calculator page has both; for the metric side of circles, the circle calculator guide picks up where this one stops. The treatment here follows OpenStax’s Intermediate Algebra and MathWorld’s Circle entry, both worth a bookmark if you want the formal statements.