Circle Theorems: past paper questions & worked solutions

\(\angle BDE=\angle ABE=52^\circ\), by the alternate segment theorem for tangent \(AB\) and chord \(BE\).

The angle between a tangent and chord equals the angle in the opposite segment.

Since \(ED=BD\), the base angles \(\angle BED\) and \(\angle EBD\) are equal. Thus \[\angle BED=\frac{180^\circ-52^\circ}{2}=64^\circ.\]

Equal sides in triangle EBD subtend equal opposite angles, and its angles sum to 180 degrees.

The angle between the tangent \(CN\) and the chord \(CB\) is \(78^\circ\).

By the alternate segment theorem,

Also,

so triangle \(ABC\) is isosceles and the base angles are equal:

Using the angle sum of a triangle:

Use the intersecting chords theorem: \[ AP\times PB=CP\times PD \]

Substitute the values: \[ 9\times 6=8\times PD \]

Since \(OA=OB\), triangle \(AOB\) is isosceles: \[ \angle ABO=36^\circ \]

So \[ \angle AOB=180^\circ-36^\circ-36^\circ=108^\circ \]

Since \(OB=OC\), triangle \(BOC\) is isosceles: \[ \angle OBC=21^\circ \]

So \[ \angle BOC=180^\circ-21^\circ-21^\circ=138^\circ \]

Angles around \(O\) add to \(360^\circ\): \[ \angle AOC=360^\circ-108^\circ-138^\circ=114^\circ \]

Since \(OA=OC\), \[ \angle ACO=\frac{180^\circ-114^\circ}{2}=33^\circ \]

A, B, C on circle centre O. ∠*B**A**O* = 36°, ∠*B**C**O* = 21°. Find ∠*A**C**O*.

△*O**A**B* is isosceles (*O**A* = *O**B*= radius): ∠*O**A**B* = ∠*O**B**A* = 36° ⟹ ∠*A**O**B* = 180° − 36° − 36° = 108°

△*O**B**C* is isosceles (*O**B* = *O**C*): ∠*O**B**C* = ∠*O**C**B* = 21° ⟹ ∠*B**O**C* = 180° − 21° − 21° = 138°

Angles at centre O (on the same side): ∠*A**O**C* = 360° − 108° − 138° = 114°

△*O**A**C* is isosceles (*O**A* = *O**C*): $$\\angle OAC + \\angle OCA = 180°-114° = 66° \\implies \\angle ACO = \\frac{66°}{2} = 33°$$

DB is a diameter; chords AC and DB intersect at P. AP = 12 cm, PC = 7 cm, PB = 6 cm. Find area of circle (3 s.f.).

Apply the **Intersecting Chords Theorem**: *A**P* × *P**C* = *B**P* × *P**D* $$12\\times7 = 6\\times PD \\implies PD = \\frac{84}{6} = 14\\text{ cm}$$

Diameter *D**B*: *D**B* = *P**D* + *P**B* = 14 + 6 = 20 cm ⟹ *r* = 10 cm

Area of circle: Area = *π**r*² = 100*π* ≈ 314.159… cm²

Intersecting Chords Theorem: if two chords cross inside a circle at P, then *A**P* × *P**C* = *B**P* × *P**D*.

Since \(OA=OC\), triangle \(AOC\) is isosceles.

So \[ \angle AOC=180^\circ-52^\circ-52^\circ=76^\circ \]

The angle at the circumference is half the angle at the centre standing on the same arc: \[ \angle ABC=\frac{76^\circ}{2}=38^\circ \]

A radius is perpendicular to a tangent, so \[ \angle OAD=90^\circ \]

and \[ \angle OCD=90^\circ \]

In quadrilateral \(AOCD\), \[ \angle AOC=360^\circ-90^\circ-90^\circ-48^\circ=132^\circ \]

The question asks for the reflex angle: \[ 360^\circ-132^\circ=228^\circ \]

From the external point \(P\), two secants meet the circle.

For the upper secant, \[ PB=14,\qquad PA=14+16=30 \]

For the lower secant, \[ PC=12,\qquad PD=12+DC \]

Using the secant theorem: \[ PB\times PA=PC\times PD \]

Opposite angles in a cyclic quadrilateral add to \(180^\circ\): \[ \angle PSR=180^\circ-136^\circ=44^\circ \]

Since \(PS\) is a diameter, \[ \angle PRS=90^\circ \]

In triangle \(PRS\), \[ \angle RPS=180^\circ-90^\circ-44^\circ=46^\circ \]

Use the intersecting chords theorem: \[ PT\times TQ=ST\times TR \]

Since \(PTQ\) is a diameter, \[ PQ=PT+TQ=16+3=19 \]

So the radius is \[ 19\div 2=9.5 \]

The tangent at \(B\) is perpendicular to the radius \(OB\), so \[ \angle OBC=90^\circ \]

In triangle \(OBC\), \[ \angle BOC=180^\circ-90^\circ-48^\circ=42^\circ \]

Since \(O,D,C\) are on a straight line, \[ \angle BOD=42^\circ \]

\(BOK\) is a diameter, so \(BOE\) is a straight line and \[ \angle DOE=180^\circ-42^\circ=138^\circ \]

The angle at the circumference is half the angle at the centre: \[ \angle DFE=\frac{138^\circ}{2}=69^\circ \]

The central angle \(\angle POS\) is \(124^\circ\). The angle at the circumference standing on the same chord \(PS\) is half this angle.

Angles opposite each other in a cyclic quadrilateral add to \(180^\circ\), so \[ n=180-62=118 \]

Since \(QOS\) is a diameter, \[ \angle QOP=180^\circ-124^\circ=56^\circ \]

Triangle \(QOP\) is isosceles because \(OQ=OP\). Therefore \[ \angle QPO=\frac{180-56}{2}=62^\circ \]

Angles in the same segment are equal (both stand on the arc \(AD\)).

Find the gradient of the radius from the centre C=(3,1) to P=(7,6).

The tangent is perpendicular to the radius at P, so its gradient is the negative reciprocal.

Use the tangent gradient and the given point of contact.

Multiply by 5 and collect all terms on one side to obtain integer coefficients.

**Part (a)(i) Angle *D**O**B*:** ∠*D**O**B* = 2 × 48^∘ = 96^∘

(a)(i) 96^∘

**Part (a)(ii) Reason:** The angle subtended by an arc at the centre is twice the angle subtended at the circumference.

**Part (b)(i) Angle *B**C**D*:** ∠*B**C**D* = 180^∘ − 48^∘ = 132^∘

(a) The angle at the centre is twice the angle at the circumference standing on chord \(DB\): \[ \angle DOB=2\times 48^\circ=96^\circ \]

(b) Opposite angles in cyclic quadrilateral \(ABCD\) add to \(180^\circ\): \[ \angle BCD=180^\circ-48^\circ=132^\circ \]

By the alternate segment theorem,

In triangle \(ABD\),

So

Angles subtended by the same chord \(CB\) are equal, so

(a) Since \(RP\) is a diameter, \[ \angle RQP=90^\circ \]

Reason: the angle in a semicircle is \(90^\circ\).

(b) Angles \(PRQ\) and \(PSQ\) stand on the same chord \(PQ\), so they are equal.

Since \(AC\) is a diameter, \[ \angle ABC=90^\circ \]

In triangle \(ABC\), \[ \angle ACB=180^\circ-90^\circ-55^\circ=35^\circ \]

Angles in the same segment are equal, so \[ \angle ADB=\angle ACB=35^\circ \]

Since \(EOG\) is a diameter, \[ \angle EDG=90^\circ \]

In triangle \(EDG\), \[ \angle DEG=180^\circ-90^\circ-42^\circ=48^\circ \]

Angles \(DEG\) and \(DFG\) stand on the same chord \(DG\), so they are equal.

The reflex angle \(POR\) is \(238^\circ\), so the smaller angle is \[ 360^\circ-238^\circ=122^\circ \]

Since \(OP=OR\), triangle \(OPR\) is isosceles: \[ \angle OPR=\frac{180^\circ-122^\circ}{2}=29^\circ \]

By the alternate segment theorem, \[ \angle QPR=60^\circ \]

Therefore \[ \angle QPO=60^\circ-29^\circ=31^\circ \]

By the alternate segment theorem, the angle between the tangent and chord \(BD\) is equal to the angle in the opposite segment: \[ \angle BFD=39^\circ \]

So \[ \angle BFE=39^\circ+18^\circ=57^\circ \]

Opposite angles in cyclic quadrilateral \(BDEF\) add to \(180^\circ\): \[ \angle BDE=180^\circ-57^\circ=123^\circ \]

Since \(AD\) is a diameter, \[ \angle ABD=90^\circ \]

So \[ \angle ABC=90^\circ-28^\circ=62^\circ \]

Opposite angles in a cyclic quadrilateral add to \(180^\circ\): \[ \angle ADC=180^\circ-62^\circ=118^\circ \]

At \(D\), \[ \angle ADC=\angle BDA+\angle BDC \]

Angle \(ABC\) and central angle \(AOC\) stand on the same arc \(AC\).

The angle at the centre is twice the angle at the circumference: \[ \angle AOC=2\times 38^\circ=76^\circ \]

Since \(OA=OC\), triangle \(AOC\) is isosceles.

By the alternate segment theorem, the angle between tangent \(BC\) and chord \(BF\) equals the angle in the opposite segment: \[ \angle BEF=70^\circ \]

Given \[ \angle EDF=40^\circ \]

Angles \(EDF\) and \(EBF\) stand on the same chord \(EF\), so \[ \angle EBF=40^\circ \]

In triangle \(BEF\), \[ \angle EFB=180^\circ-70^\circ-40^\circ=70^\circ \]

So \[ \angle EFB=\angle FBC=70^\circ \]

These are alternate interior angles, so \[ EF\parallel ABC \]

(a) Opposite angles in cyclic quadrilateral \(MNPL\) add to \(180^\circ\): \[ \angle MLP=180^\circ-58^\circ=122^\circ \]

(b) The angle at the centre is twice the angle at the circumference standing on the same arc \(MP\): \[ \angle MOP=2\times 58^\circ=116^\circ \]

The angle at the centre is twice the angle at the circumference subtended by the same arc PR containing S.

Opposite angles of cyclic quadrilateral PQRS sum to 180 degrees.

The interior angles of quadrilateral PSRO sum to 360 degrees; its other angles are POR, ORS and PSR.

For two intersecting chords in a circle,

Substitute the given lengths:

Expand:

Collect the \(y\) terms on one side:

Factorise:

(a) The angle at the centre is twice the angle at the circumference standing on the same arc \(AD\): \[ \angle AOD=2\times 54^\circ=108^\circ \]

(b) Since \(OC=OD\), triangle \(COD\) is isosceles: \[ \angle OCD=\angle CDO=28^\circ \]

So \[ \angle COD=180^\circ-28^\circ-28^\circ=124^\circ \]

Angles around \(O\) add to \(360^\circ\): \[ \angle AOC=360^\circ-108^\circ-124^\circ=128^\circ \]

Since \(OA=OC\), \[ \angle CAO=\frac{180^\circ-128^\circ}{2}=26^\circ \]

(c) Angle \(ABC\) stands on the major arc \(AC\), so \[ \angle ABC=\frac{360^\circ-128^\circ}{2}=116^\circ \]

Opposite angles in cyclic quadrilateral \(ABCD\) add to \(180^\circ\): \[ \angle BAD=180^\circ-128^\circ=52^\circ \]

The angle at the centre is twice the angle at the circumference standing on the same arc \(BD\): \[ \angle BOD=2\times 52^\circ=104^\circ \]

Since \(OB=OD\), triangle \(OBD\) is isosceles: \[ \angle OBD=\frac{180^\circ-104^\circ}{2}=38^\circ \]

In cyclic quadrilateral \(A D C B\), opposite angles add to \(180^\circ\).

Reason: opposite angles in a cyclic quadrilateral add to \(180^\circ\).

By the alternate segment theorem,

In triangle \(ADB\),

The angle at the centre is twice the angle at the circumference standing on the same chord \(AD\):

Since \(OA=OD\), triangle \(AOD\) is isosceles:

In triangle \(ABC\), use the cosine rule: \[ AC^2=8^2+7.5^2-2(8)(7.5)\cos 98^\circ \]

Opposite angles in a cyclic quadrilateral add to \(180^\circ\), so \[ \angle ADC=180^\circ-98^\circ=82^\circ \]

In triangle \(ACD\), \[ \angle CAD=180^\circ-82^\circ-35^\circ=63^\circ \]

Use the sine rule: \[ AD=\frac{AC\sin 35^\circ}{\sin 82^\circ}=6.778\ldots \]

Perimeter: \[ 8+7.5+6.778\ldots+10.529\ldots=32.807\ldots \]