How this instrument works
A circle's equation can appear in general form, x² + y² + Dx + Ey + F = 0, with the squared terms already expanded and everything mixed together, or in standard form, (x−h)² + (y−k)² = r², which shows the center (h,k) and radius r directly. Converting from general to standard form means completing the square in x and completing the square in y separately, and the results collapse to two simple formulas: h = −D⁄2 and k = −E⁄2, with the radius following from r = √(h² + k² − F).
The center coordinates come out as simply half of each linear coefficient, negated — a direct consequence of how completing the square works: x² + Dx becomes (x + D⁄2)² − (D⁄2)², so the constant that gets added and subtracted is exactly (D⁄2)², placing the center's x-coordinate at −D⁄2.
Not every general-form equation describes a real circle. If h² + k² − F comes out negative, there's no real number whose square gives that result, meaning no actual circle satisfies the original equation — sometimes called an 'imaginary circle,' a case this calculator flags rather than returning a meaningless radius.
- Enter the coefficient of x from the general-form equation into the D field.
- Enter the coefficient of y into the E field.
- Enter the constant term into the F field.
- Read Center: x (h), Center: y (k), and Radius: the sheet completes the square in both variables automatically.
Worked example — x² + y² − 4x − 6y + 4 = 0
The general-form equation x² + y² − 4x − 6y + 4 = 0 has D = −4, E = −6, F = 4. The center is h = −(−4)⁄2 = 2, k = −(−6)⁄2 = 3, and the radius is r = √(4+9−4) = √9 = 3 — converting directly to the standard form (x−2)² + (y−3)² = 9, a circle centered at (2, 3) with radius 3.
The simpler equation x² + y² − 25 = 0 has D = 0, E = 0, F = −25 — no linear terms at all, giving a center of exactly (0, 0) and a radius of √25 = 5, matching the familiar equation of a circle centered at the origin.
Questions
What is the difference between general form and standard form for a circle?
General form, x² + y² + Dx + Ey + F = 0, has the squared terms expanded and everything mixed together. Standard form, (x−h)² + (y−k)² = r², shows the center (h,k) and radius r directly, without needing any further algebra to read them off.
How do you find the center from general form?
Take half of each linear coefficient and negate it: h = −D⁄2 for the x-coordinate and k = −E⁄2 for the y-coordinate — a direct shortcut from the completing-the-square process, without needing to write out every intermediate algebraic step.
What if the equation doesn't describe a real circle?
This happens when h² + k² − F comes out negative — no real radius squares to a negative number, so no genuine circle satisfies the original equation. This case is sometimes called an 'imaginary circle,' and this calculator flags it rather than returning a meaningless result.
Why does completing the square recover the center this way?
Expanding x² + Dx into a perfect square requires adding (D⁄2)², turning it into (x + D⁄2)² − (D⁄2)² — the value added inside that square, D⁄2, is exactly the negative of the circle's x-coordinate once the equation is rearranged into standard form.
Is this the reverse of expanding standard form into general form?
Yes — expanding (x−h)² + (y−k)² = r² produces exactly x² + y² − 2hx − 2ky + (h²+k²−r²) = 0, matching general form with D = −2h, E = −2k, and F = h²+k²−r². This calculator undoes that expansion.