Equation of a circle calculator
Standard form, general form, center and radius — from whatever you know.
What is the standard form of the equation of a circle?
(x − h)² + (y − k)² = r², where (h, k) is the center and r the radius. A circle with center (−1, 3) and radius 2 is (x + 1)² + (y − 3)² = 4 (OpenStax Intermediate Algebra 2e, Example 11.6).
How do I convert the general form to standard form?
Group the x terms and the y terms, move the constant to the right and complete the square for each. x² + y² − 4x − 6y + 4 = 0 becomes (x² − 4x + 4) + (y² − 6y + 9) = −4 + 4 + 9, that is (x − 2)² + (y − 3)² = 9: center (2, 3), radius 3 (Example 11.10). If x² and y² have the same coefficient other than 1, divide by it first: 4x² + 4y² = 64 is x² + y² = 16, radius 4.
When is the equation not a real circle?
When completing the square leaves zero or a negative number on the right. r² = 0 means only the center satisfies the equation — a single point. A negative r² means no point at all: x² + y² + 2x + 2 = 0 becomes (x + 1)² + y² = −1, which has no graph.
How do I write the equation from the center and a point?
The radius is the distance from the center to the point. Center (2, 4) and point (−2, 1): r = √(4² + 3²) = 5, so the equation is (x − 2)² + (y − 4)² = 25 (Example 11.7). With the two ends of a diameter, the center is their midpoint and the radius half the distance between them.
How do I find the circle through three points?
Its center is the same distance from all three points, so it lies where the perpendicular bisectors of two chords cross — the same as putting the three points into x² + y² + Dx + Ey + F = 0 and solving for D, E and F. The points (−1, 3), (5, 3) and (2, 6) give center (2, 3) and radius 3. Three points on one straight line have no circle through them.
Where do the formulas come from?
From OpenStax Intermediate Algebra 2e, section 11.1 “Distance and Midpoint Formulas; Circles”: the standard and general forms and completing the square. Its worked examples are part of this calculator’s automated tests.
How do I tell whether a point is on, inside or outside a circle?
Substitute the point into (x − h)² + (y − k)² and compare with r². If the result equals r², the point is on the circle; if it is less, the point is inside; if it is more, the point is outside. This is the same as comparing the distance from the center with the radius.
What is the tangent to a circle at a point?
A tangent is perpendicular to the radius at the point of contact. If the radius to the point has slope m, the tangent has slope −1 ÷ m. If the radius is horizontal, the tangent is vertical, and the other way around.
Worked example from general form
Start with x² + y² + 6x − 4y − 12 = 0. Group and complete the square: the center is (−D ÷ 2, −E ÷ 2) = (−3, 2), and r² = h² + k² − F = 9 + 4 + 12 = 25. So the standard form is (x + 3)² + (y − 2)² = 25, with radius 5. The area is π × 25 = 78.54 and the circumference is 2π × 5 = 31.42. To check a point, (2, 2): (2 + 3)² + (2 − 2)² = 25, which equals r², so (2, 2) is on the circle. At that point the radius is horizontal, so the tangent line is vertical, x = 2.
Common mistakes
- Getting the signs of the center wrong. In (x − h)², a factor (x + 3) means h = −3. The center's coordinates have the opposite signs of those written in the brackets.
- Reporting r² as the radius. The right-hand side of the equation is r². Take the square root to get the radius.
- Forgetting to divide by the leading coefficient. An equation like 2x² + 2y² − 8x = 0 must be divided by 2 first, or the completing-the-square step fails.