Two circle facts plant a right angle wherever they apply: the angle in a semicircle, and the angle between a tangent and a radius. Spot them and an angle problem becomes a triangle problem.
The angle in a semicircle is \(90^\circ\): a triangle drawn on a diameter with its third corner on the circle has a right angle at that corner. A tangent meets the radius at its point of contact at \(90^\circ\).
Fig. 1: AB is a diameter, so angle ACB = 90° (angle in a semicircle). The tangent at B is perpendicular to the radius OB. Drawn to scale
Section 01
The angle in a semicircle.
In the semicircle tab, drag point C around the circle. However you move it, the angle at C stays exactly 90° as long as AB is a diameter. Then move A or B so AB no longer passes through the centre, and watch the right angle disappear.
Theorem: the angle in a semicircle is \(90^\circ\). If AB is a diameter and C is any other point on the circle, angle ACB \(= 90^\circ\).
Reason to write: "angle in a semicircle = 90°".
Radii are equal: any triangle with two radii as sides (such as OAC) is isosceles, so its base angles are equal. Look for these whenever O is joined to points on the circle.
Worked example Using the right angle
In Fig. 1, AB is a diameter and angle CAB \(= 34^\circ\). Find angle ABC.
Angle ACB \(= 90^\circ\) (angle in a semicircle).
Angles in triangle ABC add to \(180^\circ\): angle ABC \(= 180^\circ - 90^\circ - 34^\circ\).
Angle ABC \(= 56^\circ\).
Answer: \(56^\circ\)
Worked example With an isosceles triangle
AB is a diameter of a circle with centre O, and C is on the circle. Angle BOC \(= 70^\circ\). Find angle OBC and angle BAC, giving reasons.
OB = OC (radii), so triangle OBC is isosceles and angle OBC = angle OCB.
Angle OBC \(= (180^\circ - 70^\circ) \div 2 = 55^\circ\) (angles in a triangle).
The right angle is at C, the point on the circumference, NOT at the centre and not at A or B. It only holds when AB really is a diameter, so check the chord passes through O.
Section 02
Tangent meets radius.
In the tangents tab, drag the point of contact around the circle and watch the tangent turn with it. The radius to the point of contact always meets the tangent at 90°.
Theorem: the angle between a tangent and the radius at the point of contact is \(90^\circ\).
Reason to write: "angle between tangent and radius = 90°".
Right-angled triangle: if P is outside the circle and PT is a tangent at T, then triangle OTP has its right angle at T, and OP is the hypotenuse. Use angle sums, Pythagoras or trigonometry.
Worked example Angle in the tangent triangle
PT is a tangent to a circle, centre O, at T. Angle TOP \(= 52^\circ\). Find angle OPT.
Angle OTP \(= 90^\circ\) (angle between tangent and radius).
A circle has centre O and radius 5 cm. PT is a tangent at T and PT = 12 cm. Find OP.
Angle OTP \(= 90^\circ\) (tangent and radius), so OP is the hypotenuse.
\(OP^2 = 5^2 + 12^2 = 25 + 144 = 169\).
\(OP = \sqrt{169} = 13\) cm.
Answer: 13 cm
Your turn
A circle has centre O and radius 8 cm. A point P is 17 cm from O. Find the length of the tangent from P to the circle.
The radius to the point of contact T is perpendicular to the tangent, so \(PT^2 = 17^2 - 8^2 = 289 - 64 = 225\) and PT = 15 cm.
Examiner trap
OP, the line from the centre to the outside point, is the hypotenuse, not the tangent. Subtract squares to find the tangent: \(PT^2 = OP^2 - OT^2\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
AB is a diameter of a circle and C is another point on the circle. What is angle ACB?
Quick check+10 XP
What is the angle between a tangent and the radius at the point of contact?
Quick check+10 XP
O is the centre of a circle and A and B are points on the circle. What type of triangle is OAB?
Quick check+10 XP
AB is a diameter and C is on the circle. Angle CAB \(= 40^\circ\). Find angle CBA.
Section 03
Putting it together.
The find-the-angle tab gives random circle diagrams. For each, look for a diameter, a tangent, and any pair of radii before calculating, and say each reason out loud before you type the answer.
Diameter in the diagram? Look for a triangle on it: right angle at the circumference.
Tangent in the diagram? Join the centre to the point of contact: right angle there.
Two radii? Isosceles triangle: equal base angles.
Then finish with angles in a triangle, on a straight line, or round a point.
Worked example Tangent and diameter together
AB is a diameter of a circle, centre O, and the tangent at B meets the line AC extended at D, where C is on the circle. Angle CAB \(= 28^\circ\). Find angle ADB.
Angle ABD \(= 90^\circ\) (angle between tangent and radius, since OB lies along AB).
In triangle ABD: angle ADB \(= 180^\circ - 90^\circ - 28^\circ = 62^\circ\) (angles in a triangle).
Answer: \(62^\circ\)
Your turn
In the same diagram, find angle CBD. (Hint: angle ACB is in a semicircle.)
Angle ACB \(= 90^\circ\) (angle in a semicircle), so angle ABC \(= 180 - 90 - 28 = 62^\circ\). Angle ABD \(= 90^\circ\), so angle CBD \(= 90^\circ - 62^\circ = 28^\circ\).
Examiner trap
Write the circle reason in the syllabus words. 'Because it touches the circle' does not earn the mark; 'angle between tangent and radius = 90°' does.
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
AB is a diameter of a circle and C is a point on the circle. Angle ABC \(= 27^\circ\). Find angle BAC.
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PT is a tangent to a circle, centre O, at T. OT = 7 cm and PT = 24 cm. Find OP.
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PT is a tangent to a circle, centre O, at T. Angle OPT \(= 29^\circ\). Find angle TOP.
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AB is a diameter of a circle, centre O, and C is on the circle. Angle AOC \(= 100^\circ\). Find angle BAC.
Skill unlocked
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