GCSE Circle vocabulary Practice Questions

Free GCSE Circle vocabulary practice questions with full step-by-step worked solutions. Covers circle vocabulary, classifying a line segment, naming parts of a circle, circumference. Practise exam-style problems and check your method.

circle vocabularyclassifying a line segmentnaming parts of a circlecircumferencesectors and segmentscentre
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
The diagram shows a circle with centre O(0,0)O(0,0) and radius 55 cm. The point A(3,4)A(3,4) lies on the circle. What is the correct mathematical name for the line segment OAOA?
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Worked solution

  1. Write down what the diagram gives you

    O(0, 0),r=5 cmO(0,\,0), \qquad r = 5\text{ cm}

    Every circle word is decided by where the ends of the line lie, so start from the centre O(0,0)O(0,0) and the radius r=5r = 5 cm.

  2. Look at where each end of the segment lies

    O is the centre,OA2=32+42=25=52O \text{ is the centre}, \qquad OA^2 = 3^2 + 4^2 = 25 = 5^2

    One end is the centre. The other end AA satisfies OA2=25=52OA^2 = 25 = 5^2, so OA=5OA = 5 cm and AA lies on the circle.

  3. Name the segment

    radius\text{radius}

    OAOA joins the centre OO to the point AA on the circle, so it is a radius.

Answer
radius\text{radius}
Question 2
2 markseasy
The diagram shows a circle with centre O(0,0)O(0,0) and radius 55 cm. The point P(3,4)P(3,4) lies on the circle. The straight line with equation 3x+4y=253x + 4y = 25 is a tangent to the circle at PP. Work out the size of the angle between the tangent and the radius OPOP.
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Worked solution

  1. Write down the direction of each line

    OP⃗=(3, 4),direction of the tangent=(−4, 3)\vec{OP} = (3,\, 4), \qquad \text{direction of the tangent} = (-4,\, 3)

    The radius runs from O(0,0)O(0,0) to P(3,4)P(3,4). The tangent 3x+4y=253x + 4y = 25 has direction (−4,3)(-4,3), because moving that way keeps 3x+4y3x + 4y unchanged.

  2. Work out the scalar product of the two directions

    (3)(−4)+(4)(3)=−12+12=0(3)(-4) + (4)(3) = -12 + 12 = 0

    Two directions are perpendicular exactly when their scalar product is zero. Here it is zero.

  3. Write down the angle

    90∘90^\circ

    The tangent is perpendicular to the radius at the point of contact, so the angle is 90∘90^\circ. This is true for every tangent of every circle.

Answer
90∘90^\circ
Question 3
2 marksintermediate
The circumference of a circle is 24π24\pi cm. Work out the radius of the circle, in cm.
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Worked solution

  1. Write down the formula for the circumference

    C=2πrC = 2\pi r

    The circumference of a circle is C=2πrC = 2\pi r, so the radius can be recovered from it.

  2. Put in the circumference that is given

    24π=2πr24\pi = 2\pi r

    The circumference is 24π24\pi cm, so 24π=2πr24\pi = 2\pi r.

  3. Divide both sides by pi

    24=2r24 = 2r

    Dividing by π\pi removes it from both sides — this is why the exact form is easier to work with than a decimal.

  4. Divide both sides by 2

    r=242=12r = \frac{24}{2} = 12

    The radius is 1212 cm.

  5. Check by working forwards

    2π×12=24π2\pi \times 12 = 24\pi

    A radius of 1212 cm gives a circumference of 24π24\pi cm, as required.

  6. Write down the radius

    12 cm12\text{ cm}

    The radius is 1212 cm.

Answer
12 cm12\text{ cm}
Question 4
3 markshard
The circumference of a circle is 26π26\pi cm. Work out the length of the longest chord that can be drawn in this circle, in cm.
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Worked solution

  1. Decide which chord is the longest

    longest chord=diameter\text{longest chord} = \text{diameter}

    The longest chord of a circle is the one through the centre — the diameter — so the question is really asking for the diameter.

  2. Write down the formula for the circumference

    C=2πrC = 2\pi r

    The circumference is C=2πrC = 2\pi r.

  3. Put in the circumference that is given

    26π=2πr26\pi = 2\pi r

    The circumference is 26π26\pi cm.

  4. Divide both sides by pi

    26=2r26 = 2r

    Dividing by π\pi leaves whole numbers to work with.

  5. Work out the radius

    r=262=13r = \frac{26}{2} = 13

    The radius is 1313 cm.

  6. Double the radius to get the diameter

    d=2r=2×13=26d = 2r = 2 \times 13 = 26

    The diameter, and so the longest chord, is 2626 cm.

  7. Check with the other circumference formula

    C=πd=π×26=26πC = \pi d = \pi \times 26 = 26\pi

    C=πdC = \pi d gives back 26π26\pi cm, so the diameter is right.

  8. Note the common mistake

    d≠13 cmd \neq 13\text{ cm}

    Stopping at the radius is the usual error: the question asks for the longest chord, which is the whole diameter.

  9. Note the second common mistake

    C≠dC \neq d

    The circumference is the distance round the edge, not across the circle, so it is not the answer either.

  10. Write down the longest chord

    26 cm26\text{ cm}

    The longest chord is the diameter, 2626 cm.

Answer
26 cm26\text{ cm}
Question 5
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.
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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

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