To write a circle’s equation from a geometric condition, find the centre and the radius first, then write (x − a)² + (y − b)² = r². Every condition, whether a diameter, a touching axis or equal distances, is a way of telling you those two things.
This lesson completes circle coordinate methods by reversing the direction of the earlier ones. You start with a description and build the equation, as in recovering the centre and radius read backwards.
Which conditions give the centre and radius?
| Condition | Centre | Radius |
|---|---|---|
| Diameter with ends A and B | Midpoint of AB | Half the length AB |
| Centre (a, b) touches the x-axis | (a, b) | |b| |
| Centre (a, b) touches the y-axis | (a, b) | |a| |
| Passes through two points, centre on a given line | Where the line meets the perpendicular bisector | Distance to either point |
How do you do it, step by step?
- Sketch the points and the condition.
- Find the centre using the midpoint, or by setting equal distances.
- Find r or r² as a distance from the centre to a point on the circle.
- Write the equation in bracket form.
- Check each given point in the equation.
Worked example 1: a diameter
A circle has a diameter with ends A(1, 2) and B(7, 10). Find its equation.
Step 1, centre: the midpoint is ((1 + 7) ÷ 2, (2 + 10) ÷ 2) = (4, 6).
Step 2, radius: AB = √(6² + 8²) = √100 = 10, so r = 5.
Step 3, equation: (x − 4)² + (y − 6)² = 25.
Check: A gives 9 + 16 = 25 ✓ and B gives 9 + 16 = 25 ✓. Expanded, the equation is x² + y² − 8x − 12y + 27 = 0.
Worked example 2: equal distances
A circle has its centre on the x-axis and passes through (1, 3) and (5, 1). Find its equation.
Step 1, centre: let the centre be (h, 0). Distances to both points are equal, so (1 − h)² + 3² = (5 − h)² + 1².
Step 2, expand: 1 − 2h + h² + 9 = 25 − 10h + h² + 1.
Step 3, solve: 10 − 2h = 26 − 10h, so 8h = 16 and h = 2.
Step 4, radius squared: (1 − 2)² + 3² = 10.
Step 5, equation: (x − 2)² + y² = 10.
Check: (5, 1) gives (5 − 2)² + 1 = 10 ✓.
The mistake to watch for
The common slip is treating the length of the diameter as the radius.
Mistaken answer (example 1): (x − 4)² + (y − 6)² = 100
The student used AB = 10 as r, so r² = 100. The diameter is twice the radius.
The correction is to halve the diameter before squaring. A quick check is to substitute A: if the left-hand side does not equal the right-hand side, the radius is wrong. Here 9 + 16 = 25, not 100.
Check yourself
1. A circle has centre (2, −3) and touches the y-axis. Write its equation.
Show answer
The distance from the centre to the y-axis is |2| = 2, so r = 2.
(x − 2)² + (y + 3)² = 4
2. A circle has a diameter from (−2, 1) to (4, 5). Find its equation.
Show answer
Centre: (1, 3). AB² = 6² + 4² = 52, so r = √52 ÷ 2 and r² = 13.
(x − 1)² + (y − 3)² = 13. Check: (−2, 1) gives 9 + 4 = 13. ✓
3. A circle with centre on the x-axis passes through (1, 3) and (5, 1). Confirm the equation from example 2 by using the point (1, 3).
Show answer
(1 − 2)² + 3² = 1 + 9 = 10, which equals r². So (x − 2)² + y² = 10 passes through (1, 3). ✓
Where this leads next
Now combine all five skills in the circle practice set. The line and circle intersection explorer and the non-calculator working trainer are useful for checking your results.
If you can follow the examples but hesitate over which condition gives which fact, the gap is usually planning. A teacher can work on it in online one-to-one Additional Mathematics tuition.