To find the centre and radius of a circle written in expanded form, complete the square separately for x and for y. The result has the shape (x − a)² + (y − b)² = r², where (a, b) is the centre and r is the radius.
This skill opens circle coordinate methods because every later lesson needs the centre and radius. Examination questions often give x² + y² + (terms) = 0 and ask for both straight away.
Why does completing the square work here?
The circle with centre (a, b) and radius r has equation (x − a)² + (y − b)² = r². This comes from the distance formula: a point (x, y) is on the circle when its distance from (a, b) equals r, and squaring removes the square root.
Expanding gives x² + y² − 2ax − 2by + a² + b² − r² = 0. So an expanded equation is the same circle with the brackets multiplied out. Completing the square puts the brackets back.
How do you do it, step by step?
- Check the coefficients of x² and y² are both 1. If not, divide the whole equation.
- Group the x terms together and the y terms together, and move the constant to the right if you prefer.
- Halve the coefficient of x, square it, and add it to both sides. Do the same for y.
- Write each group as a bracket squared. The sign inside the bracket matches the sign of the linear term.
- Read the centre with the opposite signs from the brackets, and read r as the square root of the right-hand side.
Worked example
Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.
Step 1, group: (x² − 6x) + (y² + 4y) = 12.
Step 2, complete each square: half of −6 is −3, and (−3)² = 9. Half of 4 is 2, and 2² = 4. Add 9 and 4 to both sides.
(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
Step 3, bracket form: (x − 3)² + (y + 2)² = 25.
Step 4, read off: the centre is (3, −2) and r² = 25, so the radius is 5.
Check: the shortcut gives centre (−g, −f) with 2g = −6 and 2f = 4, so g = −3, f = 2 and the centre is (3, −2). Also r² = 9 + 4 + 12 = 25. ✓
The mistake to watch for
Two slips appear again and again: reading the centre with the wrong signs, and quoting r² as the radius.
Mistaken answer: centre (−3, 2), radius 25
The student copied the signs from inside the brackets and forgot that (x − 3) means the centre has x-coordinate positive 3. They also stopped at 25 instead of taking the square root.
The correction is to solve “x − 3 = 0” and “y + 2 = 0” to find the centre: x = 3 and y = −2. Then always ask whether the number on the right is r² or r. After completing the square it is r².
Check yourself
1. Find the centre and radius of x² + y² + 8x − 10y + 16 = 0.
Show answer
(x² + 8x + 16) + (y² − 10y + 25) = −16 + 16 + 25, so (x + 4)² + (y − 5)² = 25.
Centre (−4, 5), radius 5.
2. Find the centre and radius of 2x² + 2y² − 4x + 8y − 8 = 0.
Show answer
Divide by 2: x² + y² − 2x + 4y − 4 = 0. Then (x − 1)² + (y + 2)² = 4 + 1 + 4 = 9.
Centre (1, −2), radius 3.
3. Explain why x² + y² + 4x − 2y + 9 = 0 is not a circle.
Show answer
Completing the square gives (x + 2)² + (y − 1)² = −9 + 4 + 1 = −4. A sum of squares cannot equal a negative number, so no real circle exists.
Where this leads next
Once reading a circle feels automatic, move on to finding intersections of a line and circle, then test the whole topic with the circle practice set. The non-calculator working trainer and the quadratic structure explorer both support the completing-the-square step.
Students who follow this example but freeze on a fresh equation are often unsure where to start rather than unable to do the algebra. That is the kind of habit we look at in online one-to-one Additional Mathematics tuition.