The Equation of a Circle
This topic covers the algebraic equations of circles on the (x, y)-plane. Mastering the two standard forms allows you to quickly determine a circle's centre and radius, or to find its equation from these properties.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- A circle is defined by its centre (a, b) and its radius, r.
- The equation (x - a)² + (y - b)² = r² makes the centre (a, b) and radius r immediately obvious. Be careful with the signs for a and b.
- The general form is x² + y² + cx + dy + e = 0. You must convert this to the standard form to find the centre and radius.
- The conversion technique is 'completing the square' for both the x-terms and the y-terms separately.
- For an equation to represent a circle, the coefficients of x² and y² must be equal and non-zero. If they are not 1, divide the entire equation by their value before proceeding.
- After completing the square, if the resulting r² term is negative, the equation does not represent a real circle. If it is zero, it represents a single point.
Diagram
Formulae
(x - a)2 + (y - b)2 = r2 To define a circle with a known centre (a, b) and radius r, or when these properties are needed from the equation.
x2 + y2 + cx + dy + e = 0 Recognise this as a circle equation. Convert it to the centre-radius form by completing the square to find the circle's properties.
Definitions
- Centre-Radius Form
- The equation (x − a)² + (y − b)² = r², where (a, b) is the centre of the circle and r is its radius. Also known as the standard form.
- General Form
- The equation x² + y² + cx + dy + e = 0. This form hides the circle's properties, which can be revealed by completing the square.
- Completing the Square
- An algebraic technique to convert a quadratic expression like x² + kx into a perfect square, (x + k/2)², by adding and subtracting (k/2)².
Worked example
The equation of a circle is given by x² + y² + 8x - 12y + 27 = 0. Determine the coordinates of its centre and its radius.
- 1
Rearrange the equation to group x and y terms:
(x² + 8x) + (y² - 12y) + 27 = 0.
- 2
Complete the square for the x-terms:
x² + 8x = (x + 4)² - 4² = (x + 4)² - 16 - 3
Complete the square for the y-terms:
y² - 12y = (y - 6)² - (-6)² = (y - 6)² - 36 - 4
Substitute these back into the rearranged equation:
(x + 4)² - 16 + (y - 6)² - 36 + 27 = 0.
- 5
Combine the constant terms and move them to the right-hand side:
(x + 4)² + (y - 6)² = 16 + 36 - 27.
- 6
Simplify the right-hand side to find r²:
(x + 4)² + (y - 6)² = 25.
- 7
Identify the centre (a, b) and radius r.
The centre is (-4, 6) and the radius is √(25) = 5.
Answer: Centre: (-4, 6), Radius: 5
Common mistakes
- ×Sign errors when reading the centre coordinates from the standard form. The centre of (x + 4)² + (y - 6)² = 25 is (-4, 6), not (4, -6).
- ×Mistaking the right-hand side of the equation for the radius. The value is r², so you must take the square root to find r. If r²=25, r=5.
- ×Arithmetic mistakes when completing the square, especially when combining the constant terms. Always double-check your addition and subtraction.
No-calculator tips
- ✓When completing the square on a term like x² + 5x, work with fractions. This becomes (x + 5/2)² - (5/2)² = (x + 5/2)² - 25/4. Avoid decimals.
- ✓If r² is not a perfect square, for example 18, leave the radius as a simplified surd. √(18) = √(9 × 2) = 3*√(2). This is exact and requires no calculation.
- ✓When moving constants across the equals sign, be meticulous with sign changes to avoid simple arithmetic errors.