Equation of a circle
Also known as: circle equation, standard form of a circle
The equation of a circle in standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the radius. Every point (x, y) on the circle sits exactly r units from the center.
The standard form of a circle's equation is (x − h)² + (y − k)² = r², where the point (h, k) is the center of the circle and r is its radius. The equation is really the Pythagorean theorem in disguise: it says that every point (x, y) on the circle lies at a distance of exactly r from the center.
Reading the equation takes care with signs. Because the formula subtracts h and k, the circle (x − 3)² + (y + 2)² = 25 has center (3, −2) — not (−3, 2) — and radius 5, since r² = 25. A circle centered at the origin simplifies to x² + y² = r².
Circles are often disguised in general form, such as x² + y² − 6x + 4y − 12 = 0. To find the center and radius, complete the square in x and in y: grouping terms gives (x − 3)² + (y + 2)² = 25, revealing the same center (3, −2) and radius 5. Recognizing when a general second-degree equation is a circle — equal coefficients on x² and y², no xy term — is a standard test skill.
The SAT, ACT, and CLT all test circle equations. The SAT includes circles in its geometry questions, often requiring completing the square; the ACT covers circles among conic sections; and the CLT tests reading the center and radius from standard form. Know the formula cold, and watch the sign flips.
Key takeaways
- Standard form is (x − h)² + (y − k)² = r², with center (h, k) and radius r.
- Watch the signs: (x − 3)² + (y + 2)² = 25 has center (3, −2) and radius 5.
- A circle centered at the origin is x² + y² = r².
- Convert general form to standard form by completing the square in x and y.
- The SAT, ACT, and CLT all test identifying a circle's center and radius from its equation.