Standard Equation of a Circle Calculator
Find (x−h)²+(y−k)²=r² from center+radius, from three points, or from general form x²+y²+Dx+Ey+F=0 (completing the square) — with graph.
Enter D, E, F for x²+y²+Dx+Ey+F=0
Worked Examples (verified)
Worked example 1: Center (2, -3), radius 5
| Input | h=2, k=-3, r=5 |
| Standard equation | (x-2)^2+(y+3)^2=25 |
| Center (h,k) | (2, -3) |
| Radius | 5.0 |
| General form | x^2+y^2-4x+6y-12=0 |
Worked example 2: Through points (0,0), (4,0), (0,3)
| Input | points (0,0), (4,0), (0,3) |
| Standard equation | (x-2)^2+(y-1.5)^2=6.25 |
| Center (h,k) | (2, 1.5) |
| Radius | 2.5 |
| General form | x^2+y^2-4x-3y=0 |
Worked example 3: General form x²+y²-4x+6y-12=0
| Input | D=-4, E=6, F=-12 |
| Standard equation | (x-2)^2+(y+3)^2=25 |
| Center (h,k) | (2, -3) |
| Radius | 5.0 |
| General form | x^2+y^2-4x+6y-12=0 |
Frequently Asked Questions
What is the standard equation of a circle?
The standard (center-radius) form is (x-h)^2+(y-k)^2=r^2, where (h,k) is the center and r is the radius. Every point (x,y) on the circle is exactly r units from (h,k).
How do you write the equation for center (2,-3) and radius 5?
Substitute h=2, k=-3, r=5 into (x-h)^2+(y-k)^2=r^2: (x-2)^2+(y+3)^2=25. Its general form is x^2+y^2-4x+6y-12=0. The calculator above produces both forms instantly.
How do you find the circle equation through three points?
Solve the linear system from the three points for the center (h,k), then r is the distance to any point. Example: (0,0), (4,0), (0,3) give center (2, 1.5) and r=2.5, so the equation is (x-2)^2+(y-1.5)^2=6.25.
How do you convert x^2+y^2+Dx+Ey+F=0 to standard form?
Complete the square in x and y. Example: x^2+y^2-4x+6y-12=0 becomes (x^2-4x+4)+(y^2+6y+9)=12+4+9, i.e. (x-2)^2+(y+3)^2=25, so the center is (2,-3) and the radius is 5.
What if the three points are collinear?
Then no circle exists: the linear system's determinant is zero, so there is no unique center. The calculator reports an error instead of a fake equation.
What does it mean when r^2 comes out negative?
There is no real circle. For example D=1, E=1, F=10 gives r^2=(D^2+E^2)/4-F=0.5-10=-9.5, which is impossible for a real radius, so x^2+y^2+x+y+10=0 has no real graph.
What is the general form of a circle equation?
x^2+y^2+Dx+Ey+F=0, with D=-2h, E=-2k and F=h^2+k^2-r^2. Example: center (2,-3), radius 5 gives x^2+y^2-4x+6y-12=0.
How is this different from circle-area, circumference-to-radius, or circumscribed-circle?
circle-area computes area from radius; circumference-to-radius is a C/D/r/A dimension hub; circumscribed-circle finds a triangle's circumcircle (R=abc/4K). This tool is the analytic-geometry workbench: standard equation from center+radius, from three points, or from general form via completing the square - plus a graph of the equation.
Related Tools
- circumference-to-radius — all-pairs hub: circumference, diameter, radius, area.
- circumscribed-circle — circumcircle of a triangle (R=abc/4K).
- circle-area — area of a circle from radius.
Free tool by NRTechStudio