The tangent, and the right angle it always makes
A tangent is a line touching a circle at exactly one point, called the point of contact. A line cutting the circle at two points is a secant instead.
The tangent theorem is short and used everywhere: the tangent at any point is perpendicular to the radius drawn to that point. The moment you see a tangent in a diagram, draw the radius to the point of contact — it produces a right angle, and the right angle produces Pythagoras.
- The tangent is perpendicular to the radius at the point of contact.
- From any external point, two tangents can be drawn, and they are equal in length.
- The line from the external point to the centre bisects the angle between the two tangents, and also bisects the angle between the two radii.
- Two circles touching each other have their centres and the point of contact on one straight line.
Start with Angle at centre and drag P around the major arc. The angle at the circumference never changes, and stays exactly half the angle at O — which is where every other theorem here comes from.
The angle at the centre, and everything that follows from it
One theorem does the heavy lifting: the angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circumference. The other results in this chapter are special cases of it, which is why they are worth deriving rather than memorising separately.
| Theorem | Statement | Why it follows |
|---|---|---|
| Angle at the centre | ∠AOB = 2∠APB | the parent theorem |
| Angles in the same segment | ∠APB = ∠AQB | both are half the same central angle |
| Angle in a semicircle | ∠APB = 90° | the central angle is a straight 180°, halved |
| Cyclic quadrilateral | opposite angles sum to 180° | the two central angles make a full 360°, each halved |
| Exterior angle | equals the interior opposite angle | follows from the previous line |
| Alternate segment | tangent–chord angle = angle in the alternate segment | from the tangent–radius right angle |
Reflex angles count
When P sits on the minor arc, the relevant angle at the centre is the reflex one, and the doubling still holds. A student who uses the non-reflex angle here gets an answer that is wrong by exactly the difference — which is why the marks in this chapter go to naming the theorem you used, not just to the number.
Working a diagram
The method is always the same. Mark every radius you can see, because they are all equal and equal radii make isosceles triangles. Look for a diameter, because it gives you a right angle immediately. Name each theorem as you use it — that sentence is worth a mark on its own in nearly every mark scheme.
A, B, C and D lie on a circle with centre O. ∠BAD = 70° and ∠ADB = 40°. Find ∠BCD and ∠BOD.
- ABCD is a cyclic quadrilateral, so
∠BAD + ∠BCD = 180°.Opposite angles of a cyclic quadrilateral are supplementary. Naming the theorem earns the mark. ∠BCD = 180° − 70° = 110°.- For ∠BOD, use the angle at the centre with ∠BAD = 70° at the circumference.A and the centre are on opposite sides of the chord BD, so this is the standard configuration.
∠BOD = 2 × 70° = 140°.Check: the reflex angle at O is 360° − 140° = 220°, which is twice 110°, the angle at C. Both halves agree.
∠BCD = 110°; ∠BOD = 140°
The alternate segment theorem
This is the one students leave until last, and it is worth ten minutes. Draw a tangent touching the circle at T and a chord TC from the point of contact. The chord splits the circle into two segments. The angle between the tangent and the chord equals the angle subtended by that chord in the other segment — the one on the far side of the chord.
The word alternate is the instruction: look across the chord, not on the same side as the angle you started with.
Before you leave this chapter
- Tangent ⟂ radius at the point of contact. Two tangents from one external point are equal.
- Angle at the centre = twice the angle at the circumference on the same arc.
- Angles in the same segment are equal; the angle in a semicircle is 90°.
- Opposite angles of a cyclic quadrilateral add to 180°, and an exterior angle equals the opposite interior one.
- Tangent–chord angle equals the angle in the alternate segment. Always name the theorem you used.
Proving the angle at the centre theorem
This proof is short, it is examinable, and it explains why every other theorem in the chapter follows from this one. It uses nothing but the fact that all radii are equal.
Let A and B be two points on the circle, O the centre, and P a point on the major arc. Join PO and extend it to a point D on the far side of the circle.
- OP and OA are both radii, so triangle OAP is isosceles and ∠OPA = ∠OAP. Call this angle x.
- The exterior angle of a triangle equals the sum of the two interior opposite angles, so ∠AOD = 2x.
- The same argument on triangle OBP gives ∠OPB = ∠OBP = y and ∠BOD = 2y.
- Adding: ∠AOB = ∠AOD + ∠BOD = 2x + 2y = 2(x + y) = 2∠APB. ∎
Everything else is a corollary
Put the chord along a diameter and the central angle is 180°, so the angle at P is 90° — the angle in a semicircle. Take two points P and Q on the same arc and both are half the same central angle, so they are equal — angles in the same segment. Take P and Q on opposite arcs and the two central angles make a full 360°, so the two circumference angles total 180° — the cyclic quadrilateral. One proof, four results.