OxfordAQA International AS and A-level Mathematics: Equation of a circle, tangents and circle geometry (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover content area PP1.1 (Circle) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International AS, Unit PSM1. For explanations, use the study guide and the revision notes. Course pages: hub and checklist. Not sure what to practise? Take a free 10-minute diagnostic. Show full working throughout.

Questions

1. A circle has centre (4, −3) and radius 5.

(a) Write down its equation. [1] (b) Expand the equation and hence show that the circle passes through the origin. [2]

2. Find the centre and radius of the circle with equation x² + y² + 12x − 6y + 20 = 0. [3]

3. Find the centre and the exact radius of the circle 3x² + 3y² − 6x + 18y − 6 = 0. [3]

4. The circle C has equation x² + y² − 4x + 10y + 4 = 0.

(a) Determine the centre and the radius of C. [2] (b) C is translated by the vector (3, −1). Find the equation of the image. [1] (c) A different translation maps C onto a circle with centre (−1, 6). Find the vector of this translation. [1]

5. The segment from A(−2, 1) to B(6, 7) is a diameter of circle C.

(a) Find the equation of the circle. [3] (b) The point P has coordinates (5, 8). Use gradients to show that angle APB is a right angle, and state what this tells you about P. [2]

6. A circle has its centre on the line x = 2. It passes through P(−3, 4) and Q(9, 0). Find the equation of the circle. [6]

7. A circle has equation x² + y² + 4x − 2y − 125 = 0.

(a) Find the centre and radius. [2] (b) Find the equation of the tangent at A(7, 8), with a, b and c whole numbers, written as ax + by = c. [3] (c) Find the equation of the normal at A. [2]

8. The circle C has equation (x − 2)² + (y + 1)² = 25. The line l has equation 3x + 4y = 27.

(a) Show that l is the tangent to C at T(5, 3). [3] (b) The line l meets the x-axis at A. Find the area of triangle ACT, where C is the centre of the circle. [2]

9. A circle has equation x² + y² − 2x − 6y − 24 = 0. The line y = x − 2 meets the circle at two points, P and Q.

(a) Show that the x-coordinates of P and Q satisfy x² − 6x − 4 = 0. [3] (b) Find the coordinates of the midpoint M of PQ. [2] (c) Find the exact length of PQ. [2]

10. The circle C has equation x² + y² − 6x − 4y − 12 = 0. The points P(−1, 5) and Q(6, 6) lie on C.

(a) Determine the centre and the radius of C. [2] (b) Find the equation of the tangent at P, with a, b and c whole numbers, written as ax + by + c = 0. [3] (c) Find the equation of the tangent at Q, in the form ax + by = c. [2] (d) The two tangents meet at R. Find the coordinates of R. [2]

11. The circle C has equation x² + y² − 8x + 4y + 11 = 0.

(a) Find the centre and radius of C. [3] (b) C is translated by the vector (−3, k), where k is a constant. The image touches the x-axis and lies above it. Find k. [2] (c) Find the coordinates of the points where the image circle meets the y-axis, giving exact values. [3]

Answers

1. (a) (x − 4)² + (y + 3)² = 25 [1] (b) Expanding: x² − 8x + 16 + y² + 6y + 9 = 25, so x² + y² − 8x + 6y = 0 [1]. There is no constant term, so (0, 0) satisfies the equation and the circle passes through the origin [1]. [3] Examiner insight: “Show that” needs the expanded equation written out and a clear statement about the origin. Writing only “it works” gives nothing to credit.

2. Complete the square on both variables: (x + 6)² − 36 + (y − 3)² − 9 + 20 = 0 [1]. So (x + 6)² + (y − 3)² = 25 [1]. Centre (−6, 3), radius 5 [1]. Examiner insight: The sign flip is the usual slip. Writing the completed-square form first lets a reader see your centre came from correct working.

3. Divide by 3: x² + y² − 2x + 6y − 2 = 0 [1]. Then (x − 1)² − 1 + (y + 3)² − 9 − 2 = 0, so (x − 1)² + (y + 3)² = 12 [1]. Centre (1, −3), radius 2√3 [1]. Examiner insight: Dividing by the common coefficient is part of the method. If you complete the square without it, the right-hand side is wrong and so is the radius.

4. (a) (x − 2)² − 4 + (y + 5)² − 25 + 4 = 0 [1]. So (x − 2)² + (y + 5)² = 25: centre (2, −5), radius 5 [1]. (b) New centre (2 + 3, −5 − 1) = (5, −6). Image: (x − 5)² + (y + 6)² = 25 [1]. (c) Vector = (−1 − 2, 6 − (−5)) = (−3, 11) [1]. [4] Examiner insight: A translation leaves the radius unchanged. State the new centre explicitly so that a sign slip in the equation is easy to spot.

5. (a) Centre is the midpoint of AB: ((−2 + 6)/2, (1 + 7)/2) = (2, 4) [1]. r² = (6 − 2)² + (7 − 4)² = 16 + 9 = 25 [1]. Equation: (x − 2)² + (y − 4)² = 25 [1]. (b) Gradient AP = (8 − 1)/(5 + 2) = 1 and gradient BP = (8 − 7)/(5 − 6) = −1 [1]. The product is −1, so AP is perpendicular to BP and angle APB = 90°. Since the angle in a semicircle is a right angle, P lies on the circle with diameter AB [1]. [5] Examiner insight: Part (b) asks you to say what the right angle tells you. Stating the conclusion in words is part of the answer, not an extra.

6. Midpoint of PQ = (3, 2) [1]. Gradient PQ = (0 − 4)/(9 + 3) = −1/3 [1]. Perpendicular gradient = 3 [1]. The centre lies on the perpendicular bisector of the chord PQ: y − 2 = 3(x − 3) [1]. Putting x = 2 gives y = −1, so the centre is (2, −1) [1]. r² = (−3 − 2)² + (4 + 1)² = 50, so the equation is (x − 2)² + (y + 1)² = 50 [1]. Examiner insight: Name the property you use (the centre sits on the perpendicular bisector of any chord). Check your centre by confirming that Q is the same distance away: 49 + 1 = 50.

7. (a) (x + 2)² − 4 + (y − 1)² − 1 − 125 = 0 [1], so (x + 2)² + (y − 1)² = 130. Centre (−2, 1), radius √130 [1]. (b) Gradient of radius from (−2, 1) to (7, 8) = 7/9 [1]. Tangent gradient = −9/7 [1]. y − 8 = −(9/7)(x − 7), so 7y − 56 = −9x + 63, giving 9x + 7y = 119 [1]. (c) The normal has gradient 7/9: y − 8 = (7/9)(x − 7) [1], which gives 7x − 9y + 23 = 0. It passes through the centre: −14 − 9 + 23 = 0 [1]. [7] Examiner insight: Part (c) has two independent routes to a check: the normal must pass through the centre, and the tangent and normal gradients must multiply to −1. Use one.

8. (a) T is on the circle: (5 − 2)² + (3 + 1)² = 9 + 16 = 25, and on l: 15 + 12 = 27 [1]. Gradient of l = −3/4. Gradient of radius CT = (3 + 1)/(5 − 2) = 4/3 [1]. The product is −1, so l is perpendicular to the radius at T, and therefore l is the tangent at T [1]. (b) When y = 0, x = 9, so A is (9, 0) [1]. The angle at T is a right angle, so area = ½ × CT × AT. CT = 5 and AT = √(16 + 9) = 5, so area = 12.5 [1]. [5] Examiner insight: To show a line is a tangent you must show both that the point is on the line and circle and that the gradients are perpendicular. Checking only one of these is incomplete.

9. (a) Replacing y by x − 2 in the circle equation: x² + (x − 2)² − 2x − 6(x − 2) − 24 = 0 [1]. Expanding: x² + x² − 4x + 4 − 2x − 6x + 12 − 24 = 0 [1], so 2x² − 12x − 8 = 0, and dividing by 2 gives x² − 6x − 4 = 0 [1]. (b) The sum of the roots is 6, so the midpoint has x = 3 [1]. Then y = 3 − 2 = 1, so M = (3, 1) [1]. (c) x = 3 ± √13, so the x-coordinates differ by 2√13 [1]. The line has gradient 1, so the y-coordinates differ by the same amount and PQ = √2 × 2√13 = 2√26 [1]. [7] Examiner insight: Check with the chord property: the centre is (1, 3), CM² = 4 + 4 = 8, and 8 + 26 = 34 equals r². Spotting that kind of agreement catches slips before you hand in.

10. (a) (x − 3)² − 9 + (y − 2)² − 4 − 12 = 0 [1], so (x − 3)² + (y − 2)² = 25. Centre (3, 2), radius 5 [1]. (b) Gradient of radius to P = (5 − 2)/(−1 − 3) = −3/4 [1]. Tangent gradient = 4/3 [1]. y − 5 = (4/3)(x + 1), so 3y − 15 = 4x + 4, giving 4x − 3y + 19 = 0 [1]. (c) Gradient of radius to Q = (6 − 2)/(6 − 3) = 4/3, so tangent gradient = −3/4 [1]. y − 6 = −(3/4)(x − 6), so 4y − 24 = −3x + 18, giving 3x + 4y = 42 [1]. (d) Solve 4x − 3y = −19 and 3x + 4y = 42. Multiplying by 4 and 3 and adding: 25x = 50, so x = 2 [1]. Then y = 9, so R = (2, 9) [1]. [3] Examiner insight: Substitute R into both tangent equations as a final check: 8 − 27 + 19 = 0 and 6 + 36 = 42. A long multi-part question rewards a quick check at the end of each part.

11. (a) (x − 4)² − 16 + (y + 2)² − 4 + 11 = 0 [1], so (x − 4)² + (y + 2)² = 9 [1]. Centre (4, −2), radius 3 [1]. (b) The image has centre (4 − 3, −2 + k) = (1, k − 2) and radius 3 [1]. It touches the x-axis from above, so the centre is 3 above the axis: k − 2 = 3, so k = 5 [1]. (c) The image is (x − 1)² + (y − 3)² = 9 [1]. At x = 0: 1 + (y − 3)² = 9, so (y − 3)² = 8 [1]. Points (0, 3 + 2√2) and (0, 3 − 2√2) [1]. [8] Examiner insight: The phrase “touches the x-axis and lies above it” fixes the sign. Without “above”, the centre could also be at height −3 and you would need to deal with both cases.

Where marks are usually lost

• Reading the centre with the wrong sign from a completed-square form.
• Giving r² as the radius, or leaving it unsquare-rooted.
• Forgetting to divide by a common coefficient before completing the square.
• Mixing up the radius gradient and the tangent gradient.
• Not checking that a given point is on the circle before building a tangent from it.
• Moving the centre the wrong way under a translation, or changing the radius.
• Not stating the geometric property used (semicircle, chord bisector, tangent perpendicular to radius) when asked to show or justify.
• Rounding surd answers to decimals when the question asks for exact values.
• Leaving a tangent equation in a form that does not match the requested format.

Next steps

• Review the revision notes.
• Re-read the study guide for any method that went wrong.
• Return to the OxfordAQA A-level Mathematics hub or print the checklist.
• Take any of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section 3.2.1 PP1.1 (Circle), published by OxfordAQA.