Circle Equation & Geometry Calculator
Solve circle equations in standard or general form. Find centre, radius, circumference, area, sector area, and arc length.
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Circle Geometry Guide
What should I know about Standard and General Form?
Standard form: (x − a)² + (y − b)² = r². Where (a, b) is centre, r is radius. Example: (x−3)² + (y+2)² = 25. Centre (3, −2), radius 5. General form: x² + y² + Dx + Ey + F = 0. Convert by completing the square. Example: x² + y² − 4x + 6y + 9 = 0. (x−2)² − 4 + (y+3)² − 9 + 9 = 0. (x−2)² + (y+3)² = 4. Centre (2, −3), radius 2. Relationships: D = −2a. E = −2b. F = a² + b² − r². The general form requires the coefficients of x² and y² to be equal (so each can complete to a square) — otherwise it's not it's not a circle but an ellipse.
What's the key thing to understand about Core Formulas?
Diameter d = 2r. Circumference C = 2πr = πd. Area A = πr². Where π ≈ 3.14159. From C find r: r = C / (2π). Example: C = 31.4 → r = 31.4 / 6.28 = 5. From A find r: r = √(A/π). Example: A = 50 → r = √(50/π) = √15.92 = 3.99. Sector of angle θ (degrees): arc length = (θ/360) × 2πr. Sector area = (θ/360) × πr². Sector of angle θ (radians): arc length = rθ. Sector area = ½r²θ. Segment of a chord: area = ½r²(θ − sin θ) where θ is in radians.
What do I need to know about Equations of Tangent Lines?
Tangent to circle x² + y² = r² at point (x₁, y₁): xx₁ + yy₁ = r². Tangent to (x−a)² + (y−b)² = r² at (x₁, y₁): (x−a)(x₁−a) + (y−b)(y₁−b) = r². Tangent length from external point: √(distance² − r²). Two circles: tangent if distance between centres equals sum (externally tangent) or absolute difference (internally tangent) of radii. Intersecting if difference < distance < sum. Power of a point P with respect to circle: |PC|² − r². Where C = centre. The power equals (tangent length)² for external points, a relationship used in various geometric proofs and constructions involving circles and tangent lines.
What should I know about Applications?
Engineering: gear ratios depend on circle circumference. Architecture: arches, domes use circular geometry. Astronomy: orbital paths approximated as circles (more accurately ellipses, but circles are first approximation). Navigation: great circles — shortest paths on Earth's surface. Statistics: circle and pi appear in normal distribution (∫e^(-x²)dx). Physics: angular velocity ω = 2π/T (period). Centripetal acceleration a = v²/r. Fourier analysis: circles in the complex plane underlie all of signal processing, since sine and cosine waves are fundamentally projections of circular motion.
What should I know about Finding a Circle from Three Points?
Any three non-collinear points determine exactly one circle passing through all of them. The standard method sets up the general equation x²+y²+Dx+Ey+F=0 and substitutes each point's coordinates to get three simultaneous linear equations in D, E, F, then solves them. For points (0,0), (4,0), (0,4): substituting (0,0) gives F=0. Substituting (4,0) gives 16+4D=0, so D=−4. Substituting (0,4) gives 16+4E=0, so E=−4. The equation is x²+y²−4x−4y=0, which completes the square to (x−2)²+(y−2)²=8 — a circle centred at (2,2) with radius √8≈2.83. This method is the algebraic equivalent of the geometric construction using perpendicular bisectors of two chords, which also locates a unique centre equidistant from all three points.
What do I need to know about Circle-Line Intersection?
To find where a line intersects a circle, substitute the line's equation into the circle's equation and solve the resulting quadratic. For the circle x²+y²=25 and line y=x−1: substitute to get x²+(x−1)²=25, which expands to 2x²−2x+1=25, then 2x²−2x−24=0, simplifying to x²−x−12=0, factoring to (x−4)(x+3)=0, giving x=4 or x=−3. Corresponding y-values: (4,3) and (−3,−4). The discriminant of the resulting quadratic tells you the intersection type before even solving fully: a positive discriminant means two intersection points (the line is a secant), zero means exactly one point (the line is tangent, touching the circle at a single point), and negative means no real intersection (the line passes entirely outside the circle). This is the same discriminant logic used to classify quadratic roots elsewhere on this site, applied to a geometric intersection problem.
What do I need to know about Determining Whether a Point Is Inside, On, or Outside a Circle?
Given a circle (x-h)² + (y-k)² = r² and a test point (x₀, y₀), you can classify the point's position just by computing d² = (x₀-h)² + (y₀-k)² and comparing it to r². If d² < r², the point lies inside the circle; if d² = r², it lies exactly on the circle; and if d² > r², it lies outside. This avoids taking an unnecessary square root, since comparing squared distances directly is both faster and avoids introducing rounding error from a square-root calculation. This test is used constantly in computer graphics and game development for simple "is this click inside this circular button" or "has this object entered this circular zone" checks, where it needs to run instantly and repeatedly against many points.
Circle Equation Form Conversion Table
A circle's equation can be written in either general or standard form, and converting between them always comes down to completing the square:
| Form | Equation | What It Tells You Directly |
|---|---|---|
| General Form | x² + y² + Dx + Ey + F = 0 | Easy to expand from a given circle, hard to visualize |
| Standard Form | (x − h)² + (y − k)² = r² | Center (h, k) and radius r, visible immediately |
| Conversion | h = −D/2, k = −E/2, r = √((D/2)² + (E/2)² − F) | Derived by completing the square on x and y separately |
If (D/2)² + (E/2)² − F comes out negative, there is no real circle — the "equation" describes an empty set, since you can't take the square root of a negative number to get a real radius.
Worked Example: Converting General Form to Standard Form
Convert x² + y² − 6x + 8y − 11 = 0 into standard form and identify the center and radius. Step 1 — group x-terms and y-terms: (x² − 6x) + (y² + 8y) = 11. Step 2 — complete the square for each: for x² − 6x, half of −6 is −3, squared is 9, so x² − 6x = (x − 3)² − 9. For y² + 8y, half of 8 is 4, squared is 16, so y² + 8y = (y + 4)² − 16. Step 3 — substitute back: (x − 3)² − 9 + (y + 4)² − 16 = 11, which rearranges to (x − 3)² + (y + 4)² = 11 + 9 + 16 = 36. So the circle has center (3, −4) and radius √36 = 6. Using the shortcut formula directly as a check: D = −6, E = 8, F = −11, so h = −D/2 = 3, k = −E/2 = −4, and r = √((−3)² + (4)² − (−11)) = √(9 + 16 + 11) = √36 = 6 — confirming the same result both ways.
Worked Example: Finding a Circle's Equation from a Diameter's Endpoints
If you know the two endpoints of a diameter, you can find the full circle equation without any other information, since the center is the midpoint of the diameter and the radius is half the diameter's length. Given diameter endpoints A(2, 1) and B(8, 9), find the circle's equation. Center (midpoint of AB): h = (2+8)/2 = 5, k = (1+9)/2 = 5. Center = (5, 5). Diameter length: d = √[(8−2)² + (9−1)²] = √[36 + 64] = √100 = 10. Radius r = d/2 = 5. Standard form equation: (x − 5)² + (y − 5)² = 25. Check: point A(2,1) should lie exactly on the circle. (2−5)² + (1−5)² = 9 + 16 = 25 ✓. Point B(8,9): (8−5)² + (9−5)² = 9 + 16 = 25 ✓ — both diameter endpoints satisfy the equation exactly, confirming the circle was constructed correctly from just two given points. This "midpoint for center, half the distance for radius" shortcut is faster than the three-points method whenever you specifically know a diameter rather than three arbitrary points on the circumference, since it skips solving any system of equations entirely. It's also the exact method used in CAD software and geometry apps whenever a circle is drawn by dragging from one edge point to the opposite edge point — the software is simply computing the midpoint and half-distance behind the scenes in real time as you drag, updating the center and radius continuously as the cursor moves.