General Equation of a Circle
Find a circle's center and radius by completing the square. Learn when a general equation gives a circle, one point, or no real points.
Syllabus path: Geometry and Algebra › Plane Analytic Geometry › Equations and Properties of Circles
CSCA Exam Prep
This formula is aligned to the CSCA undergraduate admissions exam syllabus. Continue with the linked tutorial and topic practice.
Practice related questions (11) · Start mock exam · Sign up — AI explanations · Membership plans
CSCA Mathematics Formula Reference
Formula and explanation
The general equation of a circle: complete the square first
Direct answer: For real coefficients, x²+y²+Dx+Ey+F=0 becomes (x+D/2)²+(y+E/2)²=(D²+E²−4F)/4. If the right-hand side is positive, the center is (−D/2,−E/2) and the radius is its positive square root. A zero right-hand side gives one point; a negative value gives no real points.
Four steps to the center and radius
- Check that x² and y² have the same nonzero coefficient and that there is no xy term. If the common coefficient is A, divide the entire equation by A before identifying D, E and F.
- Group x²+Dx and y²+Ey. Complete each square by adding (D/2)² and (E/2)², making the same additions on the other side.
- Write (x−h)²+(y−k)²=K. A circle with radius √K exists only when K>0. Treat K=0 and K<0 separately.
- Read the center as (h,k), checking the signs inside the brackets. Expand your result back out or substitute a point on the circle to check it.
Worked example: 2x²+2y²+16x−12y=0
Divide every term by 2 to get x²+y²+8x−6y=0. Completing the square gives (x+4)²−16+(y−3)²−9=0, so (x+4)²+(y−3)²=25. The center is (−4,3) and the radius is 5. The point (1,3) is 5 units from the center and satisfies the original equation. Now vary only the constant F in the normalized general form.
| F in x²+y²+8x−6y+F=0 | Right-hand side K=25−F | Set of real points |
|---|---|---|
| 0 | 25 | Circle: center (−4,3), radius 5 |
| 25 | 0 | Only the point (−4,3) |
| 30 | −5 | No real points |
Seeing x²+y² is not enough to establish that the equation represents a circle. Normalize the coefficients and check that the completed-square right-hand side is strictly positive.
Common mistakes and limiting cases
- Reading the center's x-coordinate as 4 from (x+4)². Compare with (x−h)²: h=−4.
- Dividing only the squared terms by their common coefficient. The linear and constant terms must be divided too.
- Calling 25 the radius. It is the radius squared; the radius is the positive value 5.
- Treating K=0 as a circle with positive radius, or taking a real square root when K<0. The first case is a single point; the second has no real solution.
- Applying this formula directly to 2x²+y²−1=0. Its squared terms have different coefficients, so the curve needs a separate analysis.
Method reference and example notes
Last reviewed: . The reference explains completing the square. The numerical example and parameter comparison are AICSCA practice material, not official exam questions.
Related tutorial and examples
Equations and Properties of Circles
Continue with equations and properties of circles to build equations from geometric conditions and use point-to-line distance to study line–circle intersections.
Plan your next CSCA study step
Use this topic in a focused practice set, then connect it to the exam requirements and timed papers.