What Is the Center of a Circle Calculator?
This tool finds the center and radius of a circle given its equation in general form: \(x^2 + y^2 + Dx + Ey + F = 0\). Instead of manually completing the square, just enter the three coefficients D, E, and F to get the center coordinates \((h, k)\) and the radius instantly. It is a universal mathematics tool that works for any valid circle equation.
How to Use It
Rearrange your equation into the form \(x^2 + y^2 + Dx + Ey + F = 0\) so that the coefficients of \(x^2\) and \(y^2\) are both 1. Then read off:
- D — the number multiplying x
- E — the number multiplying y
- F — the constant term
Enter each value (including its sign) and the calculator returns the center and radius.
The Formula Explained
By completing the square, the general equation can be rewritten as \((x - h)^2 + (y - k)^2 = r^2\). This reveals that the center is at:
$$\left(h,\, k\right) = \left(-\frac{\text{D}}{2},\; -\frac{\text{E}}{2}\right)$$
The radius comes from $$r = \sqrt{\left(\frac{\text{D}}{2}\right)^2 + \left(\frac{\text{E}}{2}\right)^2 - F}.$$ If the value under the square root is negative, the equation does not describe a real circle.
Worked Example
Take \(x^2 + y^2 - 6x + 8y + 9 = 0\), so \(D = -6\), \(E = 8\), \(F = 9\).
$$\text{Center } x = -\frac{-6}{2} = 3.$$ $$\text{Center } y = -\frac{8}{2} = -4.$$ So the center is \((3, -4)\).
$$\text{Radius} = \sqrt{(-3)^2 + (4)^2 - 9} = \sqrt{9 + 16 - 9} = \sqrt{16} = 4.$$
FAQ
What if my equation has a coefficient like \(2x^2 + 2y^2\)? Divide the entire equation by that coefficient first so \(x^2\) and \(y^2\) each have a coefficient of 1, then read off D, E, and F.
Why is my radius zero? If \(\left(\frac{\text{D}}{2}\right)^2 + \left(\frac{\text{E}}{2}\right)^2 - F\) is zero or negative, the equation represents a single point or no real circle; this calculator shows 0 in that case.
What does \((h, k)\) mean? It is the standard notation for the center of a circle, where h is the x-coordinate and k is the y-coordinate.
