01 · Right angle to circle
A(0,0), B(6,0), D(6,8) form a triangle. Find its circumcircle.
Hint
The right angle is at B; AD is a diameter.
Worked solution
Centre = (3,4)
r² = 3² + 4² = 25
(x − 3)² + (y − 4)² = 25
Understand · explore · practise
Circles and triangles
Use the angle in a semicircle, recognise a diameter and find a circle through three points. Construct a circumcentre with perpendicular bisectors and verify geometric proofs.
Before you startPythagoras, perpendicular bisectors and circle equations
01 / Semicircles
A diameter subtends a right angle at the circumference.
If AB is a diameter and P is any other point on the circle, angle APB is 90°. The right angle is at P, not at the centre or either diameter endpoint.
The converse is useful too: if angle APB is a right angle, then P lies on the circle with diameter AB. In a right-angled triangle, its hypotenuse is therefore a diameter of its circumcircle.
Move P to either side of the diameter. The endpoints A and B are excluded: placing P at either one would collapse the triangle.
Keep AB as the diameterEqual axis scales
A(−4,1) and B(6,1) are diameter endpoints. P(4,5) is on the circle. Angle APB is a right angle.
02 / Find a diameter
Prove the right angle before choosing the hypotenuse.
You can prove perpendicularity using gradients or prove a right angle using squared side lengths. Identify which side is opposite that angle.
From a triangle to its circleWorked example
A = (−2,1), B = (4,3), D = (3,6)
At B, BA has gradient 1/3 and BD has gradient −3.
(1/3)(−3) = −1
Angle ABD = 90°
Therefore AD, opposite the right angle, is a diameter.
Centre = midpoint of AD = (1/2,7/2)
AD² = 5² + 5² = 50
The radius squared is 50/4 = 25/2.
(x − 1/2)² + (y − 7/2)² = 25/2
Check B: (7/2)² + (−1/2)² = 25/2.
03 / Three points
Two chord bisectors locate the centre.
Three distinct non-collinear points determine one circle. Its centre must lie on the perpendicular bisector of every chord, so intersect two bisectors, then calculate the radius using any given point.
A(−2,0), B(4,0), D(0,4)
Bisector of AB: x = 1
Bisector of AD: y = −x/2 + 3/2
Centre C = (1,1)
r² = (−2 − 1)² + (0 − 1)² = 10
(x − 1)² + (y − 1)² = 10
The third bisector is y = x, which also passes through C. This centre is the circumcentre, not the orthocentre formed by the altitudes.
Three bisectors, one centreEqual axis scales
A(−2,0), B(4,0), D(0,4) lie on (x − 1)² + (y − 1)² = 10. The perpendicular bisectors x = 1, y = −x/2 + 3/2 and y = x meet at C(1,1).
Watch perpendicular bisectors locate a circle
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Verify the circle
Substitute every point used in the claim.
Once you have a candidate circle, check all three original points. To show a fourth point is on the same circle, substitute it too; an approximate diagram is not a proof.
Circle: (x − 1)² + (y − 1)² = 10
E(2,−2): (2 − 1)² + (−2 − 1)²
= 1 + 9 = 10
Thus E lies on the circle through A, B and D above.
An alternative algebraic method
Use x² + y² + Dx + Ey + F = 0 and substitute each point. For A(−2,0), B(4,0), D(0,4), the three equations give D = −2, E = −2 and F = −8. Completing squares gives the same circle. This method still requires checking the result has positive radius.
Three distinct collinear points do not lie on one finite circle: a line can intersect a circle at most twice. Repeated points do not provide three independent conditions.
The circumcentre is inside an acute triangle, at the hypotenuse midpoint for a right triangle, and outside an obtuse triangle. Do not reject an exterior centre just because of its position.
05 / Unknown vertices
A right angle can supply a missing coordinate.
When a vertex is unknown, use perpendicularity first. Once the coordinates are known, identify the hypotenuse and use its midpoint.
A circle from a right-angle conditionWorked example
A = (0,1), B = (4,3), D = (2,q)
Angle ABD is 90°
The right angle is at B.
mAB = 1/2, so mBD = −2
(q − 3)/(2 − 4) = −2
q = 7
Now D = (2,7), and AD is the hypotenuse.
Centre = midpoint of AD = (1,4)
r² = 1² + 3² = 10
Use the distance from the centre to A.
(x − 1)² + (y − 4)² = 10
The squared distances of A, B and D from the centre are all 10.
06 / Your turn
Use geometry to choose the algebra.
Decide whether a diameter is already known before constructing two bisectors.
02 · Three points
Find the circle through (0,0), (4,0) and (0,6).
Hint
The bisectors of the horizontal and vertical chords are x = 2 and y = 3.
Worked solution
Centre = (2,3), r² = 13
(x − 2)² + (y − 3)² = 13
03 · A non-right triangle
Find the circle through A(−3,0), B(3,0), D(0,3√3).
Hint
The centre lies on x = 0. Equate its squared distances to A and D.
Worked solution
Let C = (0,b)
9 + b² = (3√3 − b)²
6√3 b = 18 ⇒ b = √3
r² = 12
x² + (y − √3)² = 12
All three side lengths are 6, so the triangle is equilateral.
04 · A fourth point
Does E(4,−2) lie on the circle (x − 1)² + (y − 1)² = 18?
Hint
Compare its squared distance with 18.
Worked solution
(4 − 1)² + (−2 − 1)² = 9 + 9 = 18
Yes, E lies on the circumference.
05 · A missing coordinate
A(−1,1), B(3,3), D(1,q) satisfy angle ABD = 90°. Find q and the circumcircle.
Hint
mAB = 1/2, so mBD = −2.
Worked solution
(q − 3)/(1 − 3) = −2 ⇒ q = 7
Centre = midpoint of AD = (0,4)
r² = 1 + 9 = 10
x² + (y − 4)² = 10
06 · A square and its circle
A(−2,0), B(2,0), D(2,4), E(−2,4) are successive vertices. Show they form a square and find its circumcircle.
Hint
All sides have length 4 and adjacent sides are horizontal/vertical.
Worked solution
The four equal sides and four right angles give a square of area 16. Its diagonals meet at (0,2), and the squared distance to any vertex is 8.
x² + (y − 2)² = 8
07 · A rotated square
A square has adjacent vertices A(1,−2), B(5,1). D lies on the side of AB reached by rotating the direction from A to B anticlockwise through 90°. Find C and D, the square’s area, and its inscribed and circumscribed circle equations.
Hint
AB changes by (4,3). An equal perpendicular change is (−3,4). The circles share the diagonal midpoint.
Worked solution
D = (−2,2), C = (2,5)
Side² = 4² + 3² = 25; area = 25
Centre = midpoint of AC = (3/2,3/2)
Inscribed: (x − 3/2)² + (y − 3/2)² = 25/4
Circumscribed: (x − 3/2)² + (y − 3/2)² = 25/2
The inside circle has radius half the side, 5/2. The circle through the vertices has radius half the diagonal, 5√2/2. The four side lines are 3x − 4y = 11, 4x + 3y = 23, 3x − 4y = −14 and 4x + 3y = −2; the centre lies equally far from each.
08 · A condition on an axis
A(−4,1), B(6,1) are diameter endpoints. A point P on the positive y-axis satisfies angle APB = 90°. Find P.
Hint
P lies on the circle with centre (1,1) and radius 5.
Worked solution
At x = 0: 1 + (y − 1)² = 25
y = 1 ± 2√6
Only 1 + 2√6 is positive, so P = (0,1 + 2√6).
07 / Recap
A right angle can reveal a diameter.
- The angle in a semicircle is at a circumference point.
- The hypotenuse of a right triangle is its circumcircle’s diameter.
- Otherwise, intersect two perpendicular bisectors.
- Three distinct non-collinear points determine one circle.
- Verify any extra point by substitution.
- Keep circumcentres and orthocentres distinct.
Section 1 of 7 · Semicircles