GCSE Equation of a circle Practice Questions

Free GCSE Equation of a circle practice questions with full step-by-step worked solutions. Covers equation of a circle, centre the origin, radius from equation, square roots. Practise exam-style problems and check your method.

equation of a circlecentre the originradius from equationsquare rootscentre of a circleorigin
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
Write down the equation of the circle with centre the origin and radius 77.
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Worked solution

  1. Recall the standard form

    x2+y2=r2x^2 + y^2 = r^2

    A circle with centre the origin (0, 0) and radius r always has this equation.

  2. Square the radius

    r=7⇒r2=72=49r = 7 \Rightarrow r^2 = 7^2 = 49

    The equation uses r squared, not r, so square the 7 first.

  3. Write the equation

    x2+y2=49x^2 + y^2 = 49

    The circle of radius 7 centred on the origin is x2+y2=49x^2 + y^2 = 49.

Answer
x2+y2=49x^2 + y^2 = 49
Question 2
2 markseasy
A circle has equation x2+y2=100x^2 + y^2 = 100. Work out the circumference of the circle. Give your answer in terms of π\pi.
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Worked solution

  1. Find the radius

    r2=100⇒r=10r^2 = 100 \Rightarrow r = 10

    Take the positive square root of 100.

  2. Use the circumference formula

    C=2πr=2π×10C = 2\pi r = 2\pi \times 10

    The circumference of any circle is 2 pi r.

  3. State the exact circumference

    C=20πC = 20\pi

    Leaving the answer as 20 pi keeps it exact rather than rounding to 62.8.

Answer
20π20\pi
Question 3
2 marksintermediate
Explain why the point (3,3)(3, 3) lies inside the circle x2+y2=25x^2 + y^2 = 25.
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Worked solution

  1. Find the radius

    r2=25⇒r=5r^2 = 25 \Rightarrow r = 5

    The circle x2+y2=25x^2 + y^2 = 25 has radius 5.

  2. Substitute the point

    32+323^2 + 3^2

    Work out x2+y2x^2 + y^2 for the point (3, 3).

  3. Evaluate

    32+32=9+9=183^2 + 3^2 = 9 + 9 = 18

    The value of x2+y2x^2 + y^2 at (3, 3) is 18.

  4. Compare with r squared

    18<2518 < 25

    x2+y2x^2 + y^2 is SMALLER than r2r^2.

  5. Interpret using distance

    18≈4.24<5\sqrt{18} \approx 4.24 < 5

    The distance from the origin to (3, 3) is sqrt(18), about 4.24, which is less than the radius 5.

  6. State the reason

    32+32=18<25⇒inside3^2 + 3^2 = 18 < 25 \Rightarrow \text{inside}

    The point is nearer to the centre than the radius, so it lies inside the circle. Comparing coordinates with the radius one at a time is NOT a valid test — (4, 4) has both coordinates below 5 but 42+42=32>254^2 + 4^2 = 32 > 25, so it is outside.

Answer
Because32+32=18,and18<25,sothepointisnearertheoriginthantheradius5Because 3^2 + 3^2 = 18, and 18 < 25, so the point is nearer the origin than the radius 5
Question 4
4 markshard
The line y=x+7y = x + 7 meets the circle x2+y2=169x^2 + y^2 = 169 at AA and BB. Work out the exact length of the chord ABAB.
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Worked solution

  1. Substitute the line into the circle

    x2+(x+7)2=169x^2 + (x + 7)^2 = 169

    Replace y by x + 7 in x2+y2=169x^2 + y^2 = 169.

  2. Expand

    (x+7)2=x2+14x+49(x + 7)^2 = x^2 + 14x + 49

    The middle term is 2 x x x 7=14x7 = 14x.

  3. Collect terms

    2x2+14x+49=1692x^2 + 14x + 49 = 169

    x2+x2=2x2x^2 + x^2 = 2x^2.

  4. Form the quadratic

    2x2+14x−120=02x^2 + 14x - 120 = 0

    Subtract 169 from both sides.

  5. Divide by 2

    x2+7x−60=0x^2 + 7x - 60 = 0

    Dividing by 2 makes it easier to factorise.

  6. Factorise

    (x+12)(x−5)=0(x + 12)(x - 5) = 0

    Two numbers with product -60 and sum +7 are +12 and -5.

  7. Solve for x

    x=−12orx=5x = -12 \quad\text{or}\quad x = 5

    Set each bracket to zero.

  8. Find the two points

    A(5, 12)andB(−12, −5)A(5,\, 12) \quad\text{and}\quad B(-12,\, -5)

    Substitute into y=x+7y = x + 7. Check: 52+122=1695^2 + 12^2 = 169 and (−12)2+(−5)2=169(-12)^2 + (-5)^2 = 169.

  9. Use the distance formula

    AB2=(5−(−12))2+(12−(−5))2=172+172=578AB^2 = (5 - (-12))^2 + (12 - (-5))^2 = 17^2 + 17^2 = 578

    The change in x and the change in y are both 17.

  10. Simplify the surd

    AB=578=289×2=172AB = \sqrt{578} = \sqrt{289 \times 2} = 17\sqrt{2}

    289=172289 = 17^2 is a perfect square, so the exact chord length is 17 sqrt(2) (about 24.04 — less than the diameter 26, as it must be).

Answer
17217\sqrt{2}
Question 5
6 markschallenging
The point (a,b)(a, b) lies on the circle x2+y2=r2x^2 + y^2 = r^2, where a≠0a \neq 0 and b≠0b \neq 0. Prove that the tangent to the circle at (a,b)(a, b) has equation ax+by=r2ax + by = r^2.
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Worked solution

  1. Set up

    x2+y2=r2,point (a, b) on the circlex^2 + y^2 = r^2, \quad \text{point } (a,\, b) \text{ on the circle}

    The centre is the origin O(0, 0) and the radius is r.

  2. Write down what is known

    a2+b2=r2a^2 + b^2 = r^2

    Because (a, b) lies on the circle, its coordinates satisfy the equation. This fact will be used at the very end.

  3. Find the gradient of the radius

    mradius=b−0a−0=bam_{\text{radius}} = \frac{b - 0}{a - 0} = \frac{b}{a}

    The radius joins O(0, 0) to (a, b). This needs a not equal to 0, which is given.

  4. State the circle property

    tangent⊥radius at the point of contact\text{tangent} \perp \text{radius at the point of contact}

    This is the key fact the whole proof rests on.

  5. Apply the perpendicular rule

    mtangent×ba=−1m_{\text{tangent}} \times \frac{b}{a} = -1

    The gradients of perpendicular lines have product -1. This needs b not equal to 0, which is also given.

  6. Find the tangent gradient

    mtangent=−abm_{\text{tangent}} = -\frac{a}{b}

    Rearranging gives the negative reciprocal of b/a.

  7. Use the point-gradient form

    y−b=−ab(x−a)y - b = -\frac{a}{b}(x - a)

    The tangent passes through (a, b) with gradient -a/b.

  8. Multiply through by b

    b(y−b)=−a(x−a)b(y - b) = -a(x - a)

    Clearing the fraction avoids messy algebra.

  9. Expand the left-hand side

    by−b2=−a(x−a)by - b^2 = -a(x - a)

    Multiply b into the bracket.

  10. Expand the right-hand side

    by−b2=−ax+a2by - b^2 = -ax + a^2

    -a times -a is +a2a^2.

  11. Collect the x and y terms

    ax+by=a2+b2ax + by = a^2 + b^2

    Add ax to both sides and add b2b^2 to both sides.

  12. Use the fact from step 2

    a2+b2=r2a^2 + b^2 = r^2

    (a, b) is on the circle, so the right-hand side is exactly r2r^2.

  13. Complete the proof

    ax+by=r2ax + by = r^2

    This is the required equation, so the result is proved.

  14. Test it

    (3, 4) on x2+y2=25⇒3x+4y=25(3,\, 4) \text{ on } x^2 + y^2 = 25 \Rightarrow 3x + 4y = 25

    Checking with a known example: the formula gives 3x+4y=253x + 4y = 25, which is the tangent found in the earlier questions.

  15. Note the special cases

    b=0⇒x=±r;a=0⇒y=±rb = 0 \Rightarrow x = \pm r; \qquad a = 0 \Rightarrow y = \pm r

    When b=0b = 0 the tangent is vertical and when a=0a = 0 it is horizontal — the gradient argument breaks down, which is exactly why the question excludes a=0a = 0 and b=0b = 0. Reassuringly, ax + by = r2r^2 still gives the right line in both cases.

Answer
Theradiusto(a,b)hasgradientb/a,sothetangenthasgradient−a/b.Usingy−b=−(a/b)(x−a)givesby−b2=−ax+a2,soax+by=a2+b2=r2,since(a,b)liesonthecircleThe radius to (a, b) has gradient b/a, so the tangent has gradient -a/b. Using y - b = -(a/b)(x - a) gives by - b^2 = -ax + a^2, so ax + by = a^2 + b^2 = r^2, since (a, b) lies on the circle

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