Hard GCSE Circle vocabulary Questions

Challenging, exam-style GCSE Circle vocabulary questions with worked solutions. Stretch yourself on the hardest circle vocabulary, classifying a line segment, tangent, position of a point problems.

circle vocabularyclassifying a line segmenttangentposition of a pointdistance from the centretangent perpendicular to the radius
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
TATA and TBTB are tangents to a circle with centre OO and radius 1010 cm, touching the circle at AA and at BB. TA=24TA = 24 cm. Work out the area of the quadrilateral OATBOATB, in cm2^2.
Show worked solution

Worked solution

  1. Use the defining property of a tangent

    OA⊥AT  ⇒  ∠OAT=90∘OA \perp AT \;\Rightarrow\; \angle OAT = 90^\circ

    A tangent is perpendicular to the radius drawn to the point of contact, so the angle at AA is a right angle.

  2. Name the two lines from the centre

    OA=OB=10 cmOA = OB = 10\text{ cm}

    OAOA and OBOB are radii, each 1010 cm.

  3. Mark the right angle at the first point of contact

    OA⊥AT⇒∠OAT=90∘OA \perp AT \Rightarrow \angle OAT = 90^\circ

    A tangent is perpendicular to the radius at the point of contact.

  4. Mark the right angle at the second point of contact

    OB⊥BT⇒∠OBT=90∘OB \perp BT \Rightarrow \angle OBT = 90^\circ

    The same is true at BB.

  5. Show the two tangents are equal

    △OAT≅△OBT (RHS)⇒BT=AT=24 cm\triangle OAT \cong \triangle OBT \ (\text{RHS}) \Rightarrow BT = AT = 24\text{ cm}

    Right angle, common hypotenuse OTOT and equal radii give congruent triangles, so the two tangents from TT are equal.

  6. Split the quadrilateral into two triangles

    OATB=△OAT+△OBTOATB = \triangle OAT + \triangle OBT

    The line OTOT cuts the kite into two congruent right-angled triangles.

  7. Write down the area of a right-angled triangle

    area=12×base×height\text{area} = \tfrac{1}{2} \times \text{base} \times \text{height}

    In triangle OATOAT the two shorter sides OAOA and ATAT meet at the right angle, so they are the base and the height.

  8. Substitute into the first triangle

    area(△OAT)=12×10×24\text{area}(\triangle OAT) = \tfrac{1}{2} \times 10 \times 24

    The legs are 1010 cm and 2424 cm.

  9. Work out that area

    12×10×24=120\tfrac{1}{2} \times 10 \times 24 = 120

    Each triangle has area 120120 cm2^2.

  10. Double it for the second triangle

    area(OATB)=2×12×10×24=10×24\text{area}(OATB) = 2 \times \tfrac{1}{2} \times 10 \times 24 = 10 \times 24

    The two triangles are congruent, so the kite is twice one of them — and the halves cancel.

  11. Work out the total area

    10×24=24010 \times 24 = 240

    The area of OATBOATB is 240240 cm2^2.

  12. Find OT as a check

    OT2=102+242=676⇒OT=26 cmOT^2 = 10^2 + 24^2 = 676 \Rightarrow OT = 26\text{ cm}

    Pythagoras gives OT=26OT = 26 cm, so the figure is consistent.

  13. Check with the kite area formula

    12×OT×AB=12×26×2×10×2426=240\tfrac{1}{2} \times OT \times AB = \tfrac{1}{2} \times 26 \times \frac{2 \times 10 \times 24}{26} = 240

    A kite's area is half the product of its diagonals, OTOT and ABAB, and that gives the same answer.

  14. Note the common mistake

    area≠120 cm2\text{area} \neq 120\text{ cm}^2

    Working out one triangle and stopping halves the answer; the kite is made of two of them.

  15. Write down the area

    240 cm2240\text{ cm}^2

    The area of OATBOATB is 240240 cm2^2.

Answer
240 cm2240\text{ cm}^2
Question 2
6 markschallenging
TATA and TBTB are tangents to a circle with centre OO and radius 2020 cm, touching the circle at AA and at BB. OT=29OT = 29 cm. Work out the area of the quadrilateral OATBOATB, in cm2^2.
Show worked solution

Worked solution

  1. Use the defining property of a tangent

    OA⊥AT  ⇒  ∠OAT=90∘OA \perp AT \;\Rightarrow\; \angle OAT = 90^\circ

    A tangent is perpendicular to the radius drawn to the point of contact, so the angle at AA is a right angle.

  2. Name the two lines from the centre

    OA=OB=20 cmOA = OB = 20\text{ cm}

    OAOA and OBOB are radii, each 2020 cm.

  3. Mark the right angle at the first point of contact

    OA⊥AT⇒∠OAT=90∘OA \perp AT \Rightarrow \angle OAT = 90^\circ

    A tangent is perpendicular to the radius at the point of contact.

  4. Mark the right angle at the second point of contact

    OB⊥BT⇒∠OBT=90∘OB \perp BT \Rightarrow \angle OBT = 90^\circ

    The same is true at BB.

  5. Show the two tangents are equal

    △OAT≅△OBT (RHS)⇒BT=AT=21 cm\triangle OAT \cong \triangle OBT \ (\text{RHS}) \Rightarrow BT = AT = 21\text{ cm}

    Right angle, common hypotenuse OTOT and equal radii give congruent triangles, so the two tangents from TT are equal.

  6. Split the quadrilateral into two triangles

    OATB=△OAT+△OBTOATB = \triangle OAT + \triangle OBT

    The line OTOT cuts the kite into two congruent right-angled triangles.

  7. Write down the area of a right-angled triangle

    area=12×base×height\text{area} = \tfrac{1}{2} \times \text{base} \times \text{height}

    In triangle OATOAT the two shorter sides OAOA and ATAT meet at the right angle, so they are the base and the height.

  8. Substitute into the first triangle

    area(△OAT)=12×20×21\text{area}(\triangle OAT) = \tfrac{1}{2} \times 20 \times 21

    The legs are 2020 cm and 2121 cm.

  9. Work out that area

    12×20×21=210\tfrac{1}{2} \times 20 \times 21 = 210

    Each triangle has area 210210 cm2^2.

  10. Double it for the second triangle

    area(OATB)=2×12×20×21=20×21\text{area}(OATB) = 2 \times \tfrac{1}{2} \times 20 \times 21 = 20 \times 21

    The two triangles are congruent, so the kite is twice one of them — and the halves cancel.

  11. Work out the total area

    20×21=42020 \times 21 = 420

    The area of OATBOATB is 420420 cm2^2.

  12. Find OT as a check

    OT2=202+212=841⇒OT=29 cmOT^2 = 20^2 + 21^2 = 841 \Rightarrow OT = 29\text{ cm}

    Pythagoras gives OT=29OT = 29 cm, so the figure is consistent.

  13. Check with the kite area formula

    12×OT×AB=12×29×2×20×2129=420\tfrac{1}{2} \times OT \times AB = \tfrac{1}{2} \times 29 \times \frac{2 \times 20 \times 21}{29} = 420

    A kite's area is half the product of its diagonals, OTOT and ABAB, and that gives the same answer.

  14. Note the common mistake

    area≠210 cm2\text{area} \neq 210\text{ cm}^2

    Working out one triangle and stopping halves the answer; the kite is made of two of them.

  15. Write down the area

    420 cm2420\text{ cm}^2

    The area of OATBOATB is 420420 cm2^2.

Answer
420 cm2420\text{ cm}^2
Question 3
5 markschallenging
TATA and TBTB are tangents to a circle with centre OO and radius 66 cm, touching the circle at AA and at BB. OT=10OT = 10 cm. Work out the area of the quadrilateral OATBOATB, in cm2^2.
Show worked solution

Worked solution

  1. Use the defining property of a tangent

    OA⊥AT  ⇒  ∠OAT=90∘OA \perp AT \;\Rightarrow\; \angle OAT = 90^\circ

    A tangent is perpendicular to the radius drawn to the point of contact, so the angle at AA is a right angle.

  2. Name the two lines from the centre

    OA=OB=6 cmOA = OB = 6\text{ cm}

    OAOA and OBOB are radii, each 66 cm.

  3. Mark the right angle at the first point of contact

    OA⊥AT⇒∠OAT=90∘OA \perp AT \Rightarrow \angle OAT = 90^\circ

    A tangent is perpendicular to the radius at the point of contact.

  4. Mark the right angle at the second point of contact

    OB⊥BT⇒∠OBT=90∘OB \perp BT \Rightarrow \angle OBT = 90^\circ

    The same is true at BB.

  5. Show the two tangents are equal

    △OAT≅△OBT (RHS)⇒BT=AT=8 cm\triangle OAT \cong \triangle OBT \ (\text{RHS}) \Rightarrow BT = AT = 8\text{ cm}

    Right angle, common hypotenuse OTOT and equal radii give congruent triangles, so the two tangents from TT are equal.

  6. Split the quadrilateral into two triangles

    OATB=△OAT+△OBTOATB = \triangle OAT + \triangle OBT

    The line OTOT cuts the kite into two congruent right-angled triangles.

  7. Write down the area of a right-angled triangle

    area=12×base×height\text{area} = \tfrac{1}{2} \times \text{base} \times \text{height}

    In triangle OATOAT the two shorter sides OAOA and ATAT meet at the right angle, so they are the base and the height.

  8. Substitute into the first triangle

    area(△OAT)=12×6×8\text{area}(\triangle OAT) = \tfrac{1}{2} \times 6 \times 8

    The legs are 66 cm and 88 cm.

  9. Work out that area

    12×6×8=24\tfrac{1}{2} \times 6 \times 8 = 24

    Each triangle has area 2424 cm2^2.

  10. Double it for the second triangle

    area(OATB)=2×12×6×8=6×8\text{area}(OATB) = 2 \times \tfrac{1}{2} \times 6 \times 8 = 6 \times 8

    The two triangles are congruent, so the kite is twice one of them — and the halves cancel.

  11. Work out the total area

    6×8=486 \times 8 = 48

    The area of OATBOATB is 4848 cm2^2.

  12. Find OT as a check

    OT2=62+82=100⇒OT=10 cmOT^2 = 6^2 + 8^2 = 100 \Rightarrow OT = 10\text{ cm}

    Pythagoras gives OT=10OT = 10 cm, so the figure is consistent.

  13. Check with the kite area formula

    12×OT×AB=12×10×2×6×810=48\tfrac{1}{2} \times OT \times AB = \tfrac{1}{2} \times 10 \times \frac{2 \times 6 \times 8}{10} = 48

    A kite's area is half the product of its diagonals, OTOT and ABAB, and that gives the same answer.

  14. Note the common mistake

    area≠24 cm2\text{area} \neq 24\text{ cm}^2

    Working out one triangle and stopping halves the answer; the kite is made of two of them.

  15. Write down the area

    48 cm248\text{ cm}^2

    The area of OATBOATB is 4848 cm2^2.

Answer
48 cm248\text{ cm}^2
Question 4
6 markschallenging
TATA and TBTB are tangents to a circle with centre OO and radius 55 cm, touching the circle at AA and at BB. TA=12TA = 12 cm. Work out the area of the quadrilateral OATBOATB, in cm2^2.
Show worked solution

Worked solution

  1. Use the defining property of a tangent

    OA⊥AT  ⇒  ∠OAT=90∘OA \perp AT \;\Rightarrow\; \angle OAT = 90^\circ

    A tangent is perpendicular to the radius drawn to the point of contact, so the angle at AA is a right angle.

  2. Name the two lines from the centre

    OA=OB=5 cmOA = OB = 5\text{ cm}

    OAOA and OBOB are radii, each 55 cm.

  3. Mark the right angle at the first point of contact

    OA⊥AT⇒∠OAT=90∘OA \perp AT \Rightarrow \angle OAT = 90^\circ

    A tangent is perpendicular to the radius at the point of contact.

  4. Mark the right angle at the second point of contact

    OB⊥BT⇒∠OBT=90∘OB \perp BT \Rightarrow \angle OBT = 90^\circ

    The same is true at BB.

  5. Show the two tangents are equal

    △OAT≅△OBT (RHS)⇒BT=AT=12 cm\triangle OAT \cong \triangle OBT \ (\text{RHS}) \Rightarrow BT = AT = 12\text{ cm}

    Right angle, common hypotenuse OTOT and equal radii give congruent triangles, so the two tangents from TT are equal.

  6. Split the quadrilateral into two triangles

    OATB=△OAT+△OBTOATB = \triangle OAT + \triangle OBT

    The line OTOT cuts the kite into two congruent right-angled triangles.

  7. Write down the area of a right-angled triangle

    area=12×base×height\text{area} = \tfrac{1}{2} \times \text{base} \times \text{height}

    In triangle OATOAT the two shorter sides OAOA and ATAT meet at the right angle, so they are the base and the height.

  8. Substitute into the first triangle

    area(△OAT)=12×5×12\text{area}(\triangle OAT) = \tfrac{1}{2} \times 5 \times 12

    The legs are 55 cm and 1212 cm.

  9. Work out that area

    12×5×12=30\tfrac{1}{2} \times 5 \times 12 = 30

    Each triangle has area 3030 cm2^2.

  10. Double it for the second triangle

    area(OATB)=2×12×5×12=5×12\text{area}(OATB) = 2 \times \tfrac{1}{2} \times 5 \times 12 = 5 \times 12

    The two triangles are congruent, so the kite is twice one of them — and the halves cancel.

  11. Work out the total area

    5×12=605 \times 12 = 60

    The area of OATBOATB is 6060 cm2^2.

  12. Find OT as a check

    OT2=52+122=169⇒OT=13 cmOT^2 = 5^2 + 12^2 = 169 \Rightarrow OT = 13\text{ cm}

    Pythagoras gives OT=13OT = 13 cm, so the figure is consistent.

  13. Check with the kite area formula

    12×OT×AB=12×13×2×5×1213=60\tfrac{1}{2} \times OT \times AB = \tfrac{1}{2} \times 13 \times \frac{2 \times 5 \times 12}{13} = 60

    A kite's area is half the product of its diagonals, OTOT and ABAB, and that gives the same answer.

  14. Note the common mistake

    area≠30 cm2\text{area} \neq 30\text{ cm}^2

    Working out one triangle and stopping halves the answer; the kite is made of two of them.

  15. Write down the area

    60 cm260\text{ cm}^2

    The area of OATBOATB is 6060 cm2^2.

Answer
60 cm260\text{ cm}^2
Question 5
5 markschallenging
TATA and TBTB are tangents to a circle with centre OO and radius 99 cm, touching the circle at AA and at BB. OT=15OT = 15 cm. Work out the perimeter of the quadrilateral OATBOATB, in cm.
Show worked solution

Worked solution

  1. Use the defining property of a tangent

    OA⊥AT  ⇒  ∠OAT=90∘OA \perp AT \;\Rightarrow\; \angle OAT = 90^\circ

    A tangent is perpendicular to the radius drawn to the point of contact, so the angle at AA is a right angle.

  2. Name the two lines from the centre

    OA=OB=9 cmOA = OB = 9\text{ cm}

    OAOA and OBOB each join the centre to a point of the circle, so both are radii of length 99 cm.

  3. Mark the right angle at the first point of contact

    OA⊥AT⇒∠OAT=90∘OA \perp AT \Rightarrow \angle OAT = 90^\circ

    The tangent TATA is perpendicular to the radius OAOA at the point of contact AA.

  4. Mark the right angle at the second point of contact

    OB⊥BT⇒∠OBT=90∘OB \perp BT \Rightarrow \angle OBT = 90^\circ

    The same property applies at BB for the tangent TBTB.

  5. Pick out the right-angled triangle OAT

    △OAT: ∠OAT=90∘, hypotenuse OT\triangle OAT:\ \angle OAT = 90^\circ,\ \text{hypotenuse } OT

    Triangle OATOAT is right-angled at AA, with OTOT as its hypotenuse.

  6. Write down Pythagoras

    OT2=OA2+AT2OT^2 = OA^2 + AT^2

    Pythagoras links the radius, the tangent and OTOT.

  7. Substitute the numbers you have

    152=92+AT215^2 = 9^2 + AT^2

    OT=15OT = 15 cm and OA=9OA = 9 cm.

  8. Rearrange for the tangent length

    AT2=225−81=144AT^2 = 225 - 81 = 144

    Subtracting gives AT2=144AT^2 = 144.

  9. Take the square root

    AT=144=12 cmAT = \sqrt{144} = 12\text{ cm}

    The first tangent is 1212 cm long.

  10. Show the second tangent has the same length

    △OAT≅△OBT (RHS)⇒BT=AT=12 cm\triangle OAT \cong \triangle OBT \ (\text{RHS}) \Rightarrow BT = AT = 12\text{ cm}

    The two triangles have a right angle, the same hypotenuse OTOT and equal radii, so they are congruent (RHS). The two tangents from an external point are therefore equal.

  11. List the four sides of the quadrilateral

    OA=9, AT=12, TB=12, BO=9 (cm)OA = 9,\ AT = 12,\ TB = 12,\ BO = 9 \ (\text{cm})

    Going round O→A→T→BO \to A \to T \to B and back to OO uses two radii and two tangents.

  12. Write down the perimeter

    perimeter=2×9+2×12\text{perimeter} = 2 \times 9 + 2 \times 12

    Two sides of length rr and two of length ATAT.

  13. Work it out

    2×9+2×12=18+24=422 \times 9 + 2 \times 12 = 18 + 24 = 42

    The perimeter is 4242 cm.

  14. Check the shape is a kite

    OA=OB,TA=TBOA = OB, \quad TA = TB

    Two pairs of adjacent equal sides — that is exactly what makes OATBOATB a kite, so the answer is consistent with the figure.

  15. Write down the perimeter

    42 cm42\text{ cm}

    The perimeter of OATBOATB is 4242 cm.

Answer
42 cm42\text{ cm}

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