A chord and a radius can determine two possible circle centres. Find the chord midpoint, calculate its perpendicular distance from the centre using Pythagoras, then measure that distance on both sides of the chord.
For a horizontal chord from (−2, 3) to (6, 3) in a circle of radius 5 units, the two possible centres are (2, 6) and (2, 0). The method combines the circle properties and coordinate geometry in the 2027 SEC G3 Mathematics K310 syllabus.
- Find the midpoint of the chord.
- Use half the chord and the radius in a right-angled triangle.
- Check both directions along the perpendicular bisector.
- Use any additional condition to select the centre.
Find the right-angled triangle
A radius drawn to either endpoint of the chord has the same length. The centre is therefore equidistant from the endpoints and lies on their perpendicular bisector.
Call the chord endpoints A and B, its midpoint M and the centre C. Triangle AMC has a right angle at M. Its hypotenuse AC is a radius, and AM is half the chord. If the radius is r and the chord length is ℓ, then CM² = r² − (ℓ/2)².
Work through the coordinate example
For A = (−2, 3) and B = (6, 3), the chord is 8 units long and its midpoint is (2, 3). The perpendicular bisector is the vertical line x = 2. In the triangle, AM = 4 units and AC = 5 units.
CM = √(5² − 4²) = 3 units. From the midpoint, move 3 units up to (2, 6) or 3 units down to (2, 0). Both centres are 5 units from each endpoint. You can verify this by checking that the horizontal and vertical differences are 4 and 3.
If the question says the centre lies above the chord, select (2, 6). If it gives no additional condition, give both centres.
Contents · Previous chapter · Next chapter
Practise with a vertical chord
A chord joins (4, −1) and (4, 11). Its midpoint is (4, 5) and its half-length is 6 units. The radius is 10 units. The centre is √(10² − 6²) = 8 units from the midpoint.
The chord is vertical, so its perpendicular bisector is horizontal. Move 8 units left or right to obtain (−4, 5) and (12, 5). If the centre has a negative x-coordinate, choose (−4, 5).
Contents · Previous chapter · Next chapter
Frequently asked questions
What if the chord is a diameter? The half-chord equals the radius, giving a midpoint-to-centre distance of zero. The midpoint is the only centre.
What if the chord is longer than twice the radius? The stated measurements cannot belong to the same circle. The chord would be longer than a diameter.
What if the chord is slanted? Find its perpendicular bisector and locate points at the calculated distance along that line. Moving straight up or down works for a horizontal chord only.