10
units
31.415927
units
78.539816
sq units
5
units
0
units
2
1=inside, 2=on, 3=outside
-0
-0
-25
-5
5
-5
5
10
units
31.415927
units
78.539816
sq units
5
units
0
units
2
1=inside, 2=on, 3=outside
-0
-0
-25
-5
5
-5
5
The Circle Equation Calculator generates the complete mathematical description of a circle from its center coordinates $$(h, k)$$ and radius $$r$$. It provides the standard form equation $$(x-h)^2 + (y-k)^2 = r^2$$, the general form $$x^2 + y^2 + Dx + Ey + F = 0$$, area, circumference, and a test to determine whether a given point lies inside, on, or outside the circle.
Circles are the simplest conic sections and the most fundamental curves in geometry. They model wheels, orbits, ripples, lenses, and countless natural phenomena. Understanding circle equations is essential for coordinate geometry, analytic geometry, and applications in physics, engineering, and computer graphics.
The standard form of a circle equation is:
$$(x - h)^2 + (y - k)^2 = r^2$$
where $$(h, k)$$ is the center and $$r$$ is the radius. Expanding this gives the general form:
$$x^2 + y^2 + Dx + Ey + F = 0$$
where $$D = -2h$$, $$E = -2k$$, and $$F = h^2 + k^2 - r^2$$.
The area enclosed by the circle is $$A = \pi r^2$$, and the circumference (perimeter) is $$C = 2\pi r$$.
For the point-on-circle test, the calculator computes the distance from the test point $$(p_x, p_y)$$ to the center:
$$d = \sqrt{(p_x - h)^2 + (p_y - k)^2}$$
If $$d < r$$, the point is inside the circle. If $$d = r$$ (within numerical tolerance), the point is on the circle. If $$d > r$$, the point is outside.
The standard form immediately reveals the center and radius, making it ideal for graphing. The general form is useful for algebraic manipulations, such as finding intersections with lines or other circles.
The test point position output uses codes: 1 for inside, 2 for on the circle, and 3 for outside. This is equivalent to checking the sign of $$(p_x - h)^2 + (p_y - k)^2 - r^2$$: negative means inside, zero means on, positive means outside.
The general form coefficients $$D$$, $$E$$, $$F$$ can be used to recover the center as $$(-D/2, -E/2)$$ and the radius as $$\sqrt{D^2/4 + E^2/4 - F}$$, which is the standard technique for completing the square.
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The point (3, 4) has distance √(9+16) = 5 = r, so it lies exactly on the circle. This is the classic 3-4-5 Pythagorean triple.
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Results
Point (2, 0) is distance 3 from center (2, −3). Since 3 < 4, the point is inside the circle. General form: x² + y² − 4x + 6y − 3 = 0.
The standard form is $$(x - h)^2 + (y - k)^2 = r^2$$, where $$(h, k)$$ is the center and $$r$$ is the radius. For a circle centered at the origin, this simplifies to $$x^2 + y^2 = r^2$$.
Given $$x^2 + y^2 + Dx + Ey + F = 0$$, complete the square: group $$(x^2 + Dx)$$ and $$(y^2 + Ey)$$, add $$(D/2)^2$$ and $$(E/2)^2$$ to both sides. The center is $$(-D/2, -E/2)$$ and radius is $$\sqrt{D^2/4 + E^2/4 - F}$$.
Compute the distance from the point to the center. If the distance equals the radius, the point is on the circle. If less, the point is inside; if greater, it is outside. Equivalently, substitute the point into $$(x-h)^2 + (y-k)^2 - r^2$$ and check the sign.
Area is $$A = \pi r^2$$ and circumference is $$C = 2\pi r$$. They are related by $$A = Cr/2$$ and $$C = 2\sqrt{\pi A}$$. Doubling the radius quadruples the area but only doubles the circumference.
No. The radius is defined as the distance from the center to any point on the circle, which is always non-negative. In the general form, if $$D^2/4 + E^2/4 - F < 0$$, the equation has no real solutions and represents an imaginary circle.
A circle is a special case of an ellipse where both semi-axes are equal ($$a = b = r$$). The eccentricity of a circle is 0, while an ellipse has $$0 < e < 1$$. Every circle equation can be written as an ellipse equation with equal denominators.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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