Equation of a Circle Calculator
Our free circle calculator solves equation acircle problems. Get worked examples, visual aids, and downloadable results.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Equation of a Circle Calculator
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Formula: (x - h)^2 + (y - k)^2 = r^2
Worked example — (x - 3)^2 + (y + 2)^2 = 25 | Area = 78.540 sq units
Formula
(x - h)^2 + (y - k)^2 = r^2
The standard form equation defines all points (x,y) at distance r from center (h,k). The general form x^2 + y^2 + Dx + Ey + F = 0 relates to the standard form by D = -2h, E = -2k, F = h^2 + k^2 - r^2.
Worked Examples
Example 1: Standard Form from Center and Radius
Problem:Write the equation of a circle with center (3, -2) and radius 5.
Solution:Center (h, k) = (3, -2), radius r = 5 Standard form: (x - 3)^2 + (y + 2)^2 = 25 General form: x^2 + y^2 - 6x + 4y - 12 = 0 D = -6, E = 4, F = -12 Circumference = 2pi(5) = 31.4159 Area = pi(25) = 78.5398
Result:(x - 3)^2 + (y + 2)^2 = 25 | Area = 78.540 sq units
Example 2: Convert General Form to Standard Form
Problem:Convert x^2 + y^2 - 6x + 4y - 12 = 0 to standard form.
Solution:D = -6, E = 4, F = -12 Center: h = -D/2 = 3, k = -E/2 = -2 r^2 = h^2 + k^2 - F = 9 + 4 + 12 = 25 r = 5 Standard form: (x - 3)^2 + (y + 2)^2 = 25
Result:Center: (3, -2) | Radius: 5 | Standard: (x-3)^2 + (y+2)^2 = 25
Frequently Asked Questions
What is the standard form equation of a circle?
The standard form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. This form directly reveals the center and radius, making it the most intuitive representation. The equation states that every point (x, y) on the circle is exactly r units away from the center (h, k), which is the geometric definition of a circle. For a circle centered at the origin, this simplifies to x^2 + y^2 = r^2. To graph a circle from standard form, plot the center point, then measure r units in all directions to find points on the circle. The standard form is particularly useful for identifying transformations: h represents a horizontal shift and k represents a vertical shift from the origin.
How do you find the equation of a circle from three points?
To find a circle through three non-collinear points (x1,y1), (x2,y2), (x3,y3), substitute each point into the general form x^2 + y^2 + Dx + Ey + F = 0 to get three linear equations in D, E, and F. For example, with points (1,1), (5,1), and (3,5): substituting gives 2 + D + E + F = 0, 26 + 5D + E + F = 0, and 34 + 3D + 5E + F = 0. Solving this system yields D, E, F, from which the center is (-D/2, -E/2) and radius is sqrt(D^2/4 + E^2/4 - F). An alternative geometric method finds the perpendicular bisectors of any two chords formed by the three points; their intersection is the center. Three collinear points do not define a circle. This construction appears in computational geometry and circumscribed circle calculations for triangles.
How do you determine if a point is inside, on, or outside a circle?
To determine a point's position relative to a circle, compute the distance from the point to the center and compare it to the radius. For point (px, py) and circle with center (h, k) and radius r, calculate d^2 = (px - h)^2 + (py - k)^2. If d^2 < r^2, the point is inside the circle. If d^2 = r^2, the point is exactly on the circle. If d^2 > r^2, the point is outside the circle. Using d^2 instead of d avoids computing a square root, which is a common optimization in computer graphics and game programming. Equivalently, substitute the point into the left side of the standard form equation: if the result is less than r^2, the point is inside. This test is fundamental to collision detection algorithms and geographic information systems.
What is the equation of a tangent line to a circle?
At a point (x1, y1) on a circle centered at origin with radius r, the tangent line equation is x1*x + y1*y = r^2. For a general circle (x-h)^2 + (y-k)^2 = r^2 with tangent point (x1, y1), the equation is (x1-h)(x-h) + (y1-k)(y-k) = r^2. The tangent line is perpendicular to the radius at the point of tangency. The slope of the radius from center (h,k) to point (x1,y1) is (y1-k)/(x1-h), so the tangent slope is -(x1-h)/(y1-k). For an external point, there are two tangent lines whose lengths equal sqrt((px-h)^2 + (py-k)^2 - r^2). Finding tangent lines is essential in optics for reflection calculations, in mechanical engineering for cam design, and in computer graphics for smooth curve rendering.
How do you find the intersection of a line and a circle?
To find where line y = mx + b intersects circle (x-h)^2 + (y-k)^2 = r^2, substitute the line equation into the circle equation: (x-h)^2 + (mx+b-k)^2 = r^2. Expanding gives a quadratic in x: (1+m^2)x^2 + 2(m(b-k)-h)x + (h^2+(b-k)^2-r^2) = 0. The discriminant determines the number of intersections: positive means two intersection points (secant line), zero means exactly one point (tangent line), and negative means no intersection (line misses the circle). For a vertical line x = c, substitute directly to get (y-k)^2 = r^2 - (c-h)^2. This calculation is fundamental to ray tracing in computer graphics, circle packing problems, and geometric construction algorithms.
What are the x-intercepts and y-intercepts of a circle?
The x-intercepts occur where y = 0, found by solving (x-h)^2 + k^2 = r^2, giving x = h plus or minus sqrt(r^2 - k^2). Real x-intercepts exist only when |k| is less than or equal to r (the circle reaches or crosses the x-axis). Similarly, y-intercepts occur where x = 0, found by solving h^2 + (y-k)^2 = r^2, giving y = k plus or minus sqrt(r^2 - h^2). Real y-intercepts exist when |h| is less than or equal to r. A circle can have 0, 1, or 2 intercepts on each axis. When |k| = r, the circle is tangent to the x-axis (one x-intercept). When the circle passes through the origin, both x = 0 and y = 0 satisfy the equation simultaneously. Intercepts are useful for graphing and for finding where circular paths cross reference axes.
How do you find the equation of a circle given center and a point on the circle?
If you know the center (h, k) and one point (x1, y1) on the circle, the radius is simply the distance between them: r = sqrt((x1-h)^2 + (y1-k)^2). Then the equation is (x-h)^2 + (y-k)^2 = (x1-h)^2 + (y1-k)^2. For example, center (2, 3) and point (5, 7): r = sqrt(9+16) = sqrt(25) = 5, so the equation is (x-2)^2 + (y-3)^2 = 25. This is the simplest way to define a unique circle because a center and one point are the minimum information needed (a center defines the location and one point defines the radius). This method is commonly used in CAD software where users click a center point and drag to a point on the circumference to create a circle.
What is completing the square and how does it apply to circle equations?
Completing the square is an algebraic technique that transforms a quadratic expression into a perfect square plus a constant. For circle equations, it converts general form to standard form. Given x^2 + y^2 - 6x + 4y - 12 = 0, group and complete: (x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4, giving (x-3)^2 + (y+2)^2 = 25. The key step is adding (coefficient/2)^2 to both sides for each variable. For x^2 - 6x, add (-6/2)^2 = 9. For y^2 + 4y, add (4/2)^2 = 4. This technique reveals that the center is (3, -2) and radius is 5. Completing the square is one of the most important algebraic manipulations in mathematics, also used for deriving the quadratic formula, analyzing conic sections, and converting quadratic forms in optimization.
How do two circles relate to each other in terms of their equations?
Two circles can be externally tangent, internally tangent, intersecting at two points, concentric, or non-intersecting. The relationship depends on the distance d between centers compared to the sum and difference of radii. If d = r1 + r2, they are externally tangent. If d = |r1 - r2|, internally tangent. If |r1 - r2| < d < r1 + r2, they intersect at two points. If d = 0, they are concentric. Otherwise they do not intersect. The radical axis (the line containing the intersection points or the locus of points with equal power) is found by subtracting one circle equation from the other, eliminating the squared terms and leaving a linear equation. Two intersection points can be found by solving the radical axis equation simultaneously with either circle equation. These relationships are fundamental to Apollonius problems, circle packing, and Venn diagram geometry.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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