Circle Theorems: Question 3

Syllabus E4.7, E4.8

Multiple choice Extended 1 mark

PAQPAQ is a straight line that is a tangent to a circle at the point AA. BB and CC are points on the circle, and CC lies on the major arc ABAB (in the alternate segment, on the opposite side of the chord ABAB from PP). The chord ABAB makes an angle of 64∘64^{\circ} with the tangent, so that angle PAB=64∘PAB = 64^{\circ}.

Work out the size of angle ACBACB.

Choose an answer to check it, then compare with the worked solution below.

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Worked solution

Identifying the correct theorem

The line PAQPAQ touches the circle at AA, so it is a tangent, and ABAB is a chord drawn from the point of contact. The angle between the tangent and the chord is ∠PAB=64∘\angle PAB = 64^{\circ}.

The alternate segment theorem states:

The angle between a tangent and a chord equals the angle in the alternate segment: the angle subtended by that chord from the arc on the other side of the chord.

Here the chord is ABAB and the angle in the alternate segment is the angle subtended by ABAB at CC, namely ∠ACB\angle ACB.

Applying the theorem

∠ACB=∠PAB=64∘.\angle ACB = \angle PAB = 64^{\circ}.

Alternative method (radius, isosceles triangle, angle at centre)

This confirms the answer using two other circle facts.

Let OO be the centre and draw radii OAOA and OBOB.

Step 1: Tangent meets radius at 90∘90^{\circ}. Since OA⊥OA \perp tangent, ∠OAB=90∘−∠PAB=90∘−64∘=26∘.\angle OAB = 90^{\circ} - \angle PAB = 90^{\circ} - 64^{\circ} = 26^{\circ}.

Step 2: Isosceles triangle. OA=OBOA = OB (both radii), so triangle OABOAB is isosceles and ∠OBA=∠OAB=26∘.\angle OBA = \angle OAB = 26^{\circ}.

Step 3: Angle sum. ∠AOB=180∘−26∘−26∘=128∘.\angle AOB = 180^{\circ} - 26^{\circ} - 26^{\circ} = 128^{\circ}.

Step 4: Angle at centre =2×= 2\times angle at circumference. The chord ABAB subtends ∠AOB\angle AOB at the centre and ∠ACB\angle ACB at the circumference (with CC on the major arc), so ∠ACB=12 ∠AOB=12(128∘)=64∘.\angle ACB = \tfrac{1}{2}\,\angle AOB = \tfrac{1}{2}(128^{\circ}) = 64^{\circ}.

Both methods agree.

Answer

∠ACB=64∘(option B)\boxed{\angle ACB = 64^{\circ}} \quad\text{(option B)}

Note how the figures 26∘26^{\circ}, 128∘128^{\circ} and 116∘116^{\circ} all appear as intermediate or mis-paired values, which is exactly why they are tempting wrong answers.