Find the equation of a circle in standard form (\((x - h)^2 + (y - k)^2 = r^2\)) and general form (\(x^2 + y^2 + Dx + Ey + F = 0\)). Solve from center and radius, 2 diameter endpoints, 3 points (circumcircle), or convert general form to standard form via step-by-step completing the square.
Select input method to generate standard and general circle equations.
Center: (3.00, -4.00) • Radius: 5.000 • Real Circle
1. Group variables: (x² - 6x) + (y² + 8y) = 11
2. Add (D/2)² & (E/2)²: (x² - 6x + 9) + (y² + 8y + 16) = 11 + 9 + 16
3. Factor perfect squares: (x - 3)² + (y + 4)² = 36 ⇒ Center (3, -4), r = 6
| 📍 Center Coordinates (\(h, k\)) | (3.000, -4.000) |
| 📏 Radius (\(r\)) | 5.000 |
| 📏 Radius Squared (\(r^2\)) | 25.000 |
| 📏 Diameter (\(d\)) | 10.000 |
| 🔄 Circumference (\(C\)) | 31.416 (2πr) |
| 🔵 Circle Area (\(A\)) | 78.540 (πr²) |
| 🎯 Circle Classification | Real Circle (\(r^2 > 0\)) |
The Cartesian equation of a circle is fundamental to analytic geometry, computer graphics rendering, robotics obstacle avoidance, and GPS satellite trilateration. Our Equation of a Circle Calculator solves critical mathematical and engineering challenges:
Finds the unique circumscribed circle equation passing through any 3 non-collinear boundary points, essential for cell tower positioning, triangular mesh generation, and radar tracking.
Converts expanded equations (\(x^2 + y^2 + Dx + Ey + F = 0\)) into center-radius standard form with step-by-step algebraic breakdown and degenerate circle detection.
Constructs circle equations directly from 2 antipodal diameter endpoints using \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\) without rounding errors.
Generates Cartesian standard form, expanded polynomial general form, and trigonometric parametric equations (\(x = h + r\cos t, y = k + r\sin t\)).
Choose from Center & Radius, Center & Point, 2 Diameter Points, 3 Points, or General Form.
Input center \((h, k)\), radius \(r\), boundary points, or general form coefficients \(D, E, F\).
Inspect standard form, general form, parametric form, center, radius, and completing-the-square steps.
Click Copy Solution Telemetry Card for math homework, CAD design, or computer graphics.
$$(x - h)^2 + (y - k)^2 = r^2$$ $$\text{Center: } (h, k), \quad \text{Radius: } r = \sqrt{r^2}$$ $$\text{Circumference: } C = 2\pi r, \quad \text{Area: } A = \pi r^2$$
$$x^2 + y^2 + Dx + Ey + F = 0$$ $$h = -\frac{D}{2}, \quad k = -\frac{E}{2}, \quad r = \sqrt{\frac{D^2 + E^2 - 4F}{4}}$$ $$\text{Real Circle: } D^2 + E^2 - 4F > 0$$
$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ $$h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}$$ $$r = \frac{1}{2}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
$$x(t) = h + r\cos(t), \quad y(t) = k + r\sin(t) \quad (0 \le t < 2\pi)$$ $$\text{Tangent at } (x_1, y_1): (x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2$$
In the standard equation \((x - h)^2 + (y - k)^2 = r^2\), the signs inside the parentheses are negative. For center \((3, -4)\), the equation is \((x - 3)^2 + (y + 4)^2 = r^2\), NOT \((x + 3)^2 + (y - 4)^2\).
The right side of the standard equation equals \(r^2\), not \(r\). For \((x - 1)^2 + (y - 2)^2 = 25\), the radius is \(r = \sqrt{25} = 5\), not \(25\).
If given \(2x^2 + 2y^2 - 8x + 12y - 6 = 0\), you must divide the entire equation by \(2\) first before completing the square to ensure \(x^2\) and \(y^2\) have coefficient \(1\).
If 3 input points lie along a straight line (collinear), their determinant is zero and no unique circle can pass through all three. Our calculator validates collinearity automatically.
Convert \(x^2 + y^2 - 6x + 8y - 11 = 0\) to standard form and find the center and radius.
Solution: Group terms: \((x^2 - 6x) + (y^2 + 8y) = 11\). Add \((-3)^2 = 9\) and \(4^2 = 16\): \((x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 = 36\). Factor: \(\mathbf{(x - 3)^2 + (y + 4)^2 = 36}\). Center: \(\mathbf{(3, -4)}\), Radius: \(\mathbf{r = 6}\).
Find the standard equation of the circle having diameter endpoints at \(A(2, 5)\) and \(B(8, -3)\).
Solution: Midpoint Center: \(h = \frac{2 + 8}{2} = 5, k = \frac{5 - 3}{2} = 1 \implies \mathbf{C(5, 1)}\). Radius: \(r = \frac{1}{2}\sqrt{(8-2)^2 + (-3-5)^2} = \frac{1}{2}\sqrt{36 + 64} = \frac{10}{2} = \mathbf{5}\). Standard Equation: \(\mathbf{(x - 5)^2 + (y - 1)^2 = 25}\).
Find the circle passing through \(A(0, 0)\), \(B(4, 0)\), and \(C(0, 6)\).
Solution: Because triangle \(ABC\) has a right angle at origin \((0, 0)\), the hypotenuse from \((4, 0)\) to \((0, 6)\) is the circle diameter! Center: \(h = \frac{4+0}{2} = 2, k = \frac{0+6}{2} = 3 \implies \mathbf{C(2, 3)}\). Radius: \(r = \sqrt{2^2 + 3^2} = \sqrt{13} \approx \mathbf{3.606}\). Standard Equation: \(\mathbf{(x - 2)^2 + (y - 3)^2 = 13}\).
Standard and general representations across canonical coordinate configurations.
| Configuration | Center (\(h, k\)) | Radius (\(r\)) | Standard Form | General Form |
|---|---|---|---|---|
| Unit Circle at Origin | \((0, 0)\) | \(1.000\) | \(x^2 + y^2 = 1\) | \(x^2 + y^2 - 1 = 0\) |
| Center \((3, -4)\), \(r = 5\) | \((3, -4)\) | \(5.000\) | \((x-3)^2 + (y+4)^2 = 25\) | \(x^2 + y^2 - 6x + 8y = 0\) |
| Diameter \((2, 5)\) & \((8, -3)\) | \((5, 1)\) | \(5.000\) | \((x-5)^2 + (y-1)^2 = 25\) | \(x^2 + y^2 - 10x - 2y + 1 = 0\) |
| 3-Points \((0,0),(4,0),(0,6)\) | \((2, 3)\) | \(3.606\) (\(\sqrt{13}\)) | \((x-2)^2 + (y-3)^2 = 13\) | \(x^2 + y^2 - 4x - 6y = 0\) |
Authoritative answers to common questions about circle equations, completing the square, 3-point circles, and coordinate geometry.
