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Circle Theorems

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 12 marks

A and B are points on a circle, centre O.

The radius of the circle is 14.5 cm.

M is the point on the chord AB such that OM is perpendicular to AB.

OM=10 cm

Work out the length of AB.

Hint

The perpendicular from the centre to a chord bisects the chord, so triangle OAM is right-angled and AM is half of AB.

Worked solution
  1. OA is a radius, so OA=14.5 cm.
  2. Triangle OAM has a right angle at M.
  3. Pythagoras: AM^2=14.5^2-10^2=210.25-100=110.25
  4. AM=\sqrt{110.25}=10.5 cm
  5. The perpendicular from the centre to a chord bisects the chord, so AB=2\times AM
  6. AB=21 cm

Question 22 marks

A, B, C and D are points on a circle, centre O.

Angle AOC=136^\circ

Work out the size of angle ADC.

Hint

First find angle ABC using the angle at the centre, then use the fact that ABCD is a cyclic quadrilateral.

Worked solution
  1. Angle AOC and angle ABC stand on the same arc AC.
  2. The angle at the centre is twice the angle at the circumference, so angle ABC=136\div 2=68^\circ
  3. ABCD is a cyclic quadrilateral, so opposite angles add up to 180^\circ.
  4. Angle ADC=180-68=112^\circ

Question 3Challenge5 marks

A, B, C and D are points on a circle.

PDT is the tangent to the circle at D.

Angle CDT=2x Angle ABD=x+10^\circ Angle ADC=5x-30^\circ

(a)

Work out the value of x.

3 marks

(b)

AB=AD

Work out the size of angle BCD.

2 marks

Hint

Angle CDT is between the tangent and the chord DC: use the alternate segment theorem to find an angle at B, then look for a pair of opposite angles in the cyclic quadrilateral.

Worked solution

Part (a)

  1. Angle DBC= angle CDT=2x
  2. (Alternate segment theorem.)
  3. So angle ABC=(x+10)+2x=3x+10
  4. Angles ABC and ADC are opposite angles of the cyclic quadrilateral ABCD, so they add up to 180^\circ.
  5. (3x+10)+(5x-30)=180
  6. 8x-20=180
  7. 8x=200, so x=25

Part (b)

  1. Angle ABD=25+10=35^\circ
  2. AB=AD, so triangle ABD is isosceles and angle ADB= angle ABD=35^\circ
  3. (Base angles of an isosceles triangle are equal.)
  4. Angle DAB=180-35-35=110^\circ
  5. (Angles in a triangle add up to 180^\circ.)
  6. Angle BCD=180-110=70^\circ
  7. (Opposite angles of a cyclic quadrilateral add up to 180^\circ.)

More on this topic: Circle Theorems worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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