OCR A Level Mathematics A: Straight lines and circles (H240) – 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 straight lines and circles in OCR A Level Mathematics A (H240), content references 1.03a to 1.03f of the Cambridge OCR specification, Version 3.1 (August 2026). All of this content is Stage 1, common to AS and A Level. The marks shown are our own guide to the size of each step. If a topic is unfamiliar, start with the study guide; the revision notes recap each method. Related links: the course hub, the checklist and the Pure Mathematics practice set.

Questions

1. Find the equation of the straight line through (6, -1) and (-2, 3), written as ax + by + c = 0 with integer a, b and c. [3]

2. The line l has equation 3x - 2y + 5 = 0 and P is the point (7, 0).

(a) Find the equation of the line that passes through P at right angles to l. [2] (b) Find the coordinates of the foot of this perpendicular on l. [2] (c) Find the shortest distance from P to l, giving an exact answer. [1]

3. A is (-4, 1) and B is (2, 5). Find the equation of the perpendicular bisector of AB in the form ax + by + c = 0. [4]

4. A pool is filled at a steady rate. It holds V cubic metres of water, with V = 14.6 after 2 hours and V = 31.4 after 6 hours.

(a) Find a linear model for V in terms of the time t hours. [3] (b) Interpret the gradient and the constant term in context. [2] (c) Find, to 3 significant figures, when the model says the pool holds 50 cubic metres. [1] (d) State one reason the model may be unreliable. [1]

5. A circle has centre (-2, 5) and passes through (3, 8).

(a) Find its equation. [2] (b) Decide whether (0, 10) and (3, 0) lie inside, on or outside the circle. [2]

6. (a) Find the centre and radius of the circle 3x² + 3y² + 12x - 18y - 3 = 0. [3] (b) The equation x² + y² + 12x - 4y + k = 0 represents a circle. Find the range of possible values of k. [2] (c) Find the value of k for which this circle has radius 5. [1]

7. Circle C1 has equation x² + y² - 2x - 4y - 4 = 0 and circle C2 has equation x² + y² - 18x - 4y + 69 = 0.

(a) Find the centre and radius of each circle. [2] (b) Show that the circles do not intersect, and find the shortest distance between a point on one and a point on the other. [2]

8. A(1, -1) and B(9, 5) are the ends of a diameter of a circle.

(a) Show that P(2, 6) lies on the circle, without using the equation of the circle. [3] (b) Find the equation of the circle. [2]

9. The line x + y = 7 meets the circle (x - 2)² + (y + 1)² = 26 at A and B.

(a) Find A and B. [3] (b) Use a circle property to show that the foot of the perpendicular from the centre to AB is the midpoint of AB. [2] (c) Find the length of AB. [1]

10. A circle has equation x² + y² - 2x - 6y - 10 = 0.

(a) Find the centre and radius. [2] (b) Show that (5, 5) lies on the circle and find the equation of the tangent there, in the form ax + by + c = 0. [3] (c) The tangent meets the axes at X and Y. Find the area of triangle OXY, where O is the origin. [2]

11. The line y = 3x + k and the circle (x - 3)² + (y - 1)² = 10 are given.

(a) Show that any point of intersection has x-coordinate satisfying 10x² + (6k - 12)x + k² - 2k = 0. [2] (b) Show that the discriminant equals -4(k - 2)(k + 18). [2] (c) Find the values of k for which the line is a tangent, and the range of k for which it meets the circle at two points. [2] (d) For the tangent with positive k, find the point of contact. [1]

Answers

1. Gradient = (3 - (-1))/(-2 - 6) = -1/2 [1]. y + 1 = -(1/2)(x - 6), so 2y + 2 = -x + 6 [1]. x + 2y - 4 = 0 [1] [3] Examiner insight: a required form is part of the question, so finish in it with integer coefficients; a correct line left as y = -x/2 + 2 leaves the last step undone. Substituting the second point as a check is quick.

2. (a) Gradient of l is 3/2, so the perpendicular gradient is -2/3 [1]. y - 0 = -(2/3)(x - 7), giving 2x + 3y - 14 = 0 [1] (b) Solve 3x - 2y = -5 with 2x + 3y = 14. Multiply by 3 and 2: 9x - 6y = -15 and 4x + 6y = 28; add: 13x = 13 [1]. x = 1, y = 4, so the foot is (1, 4) [1] (c) Distance = √[(7 - 1)² + (0 - 4)²] = √52 = 2√13 [1] [5] Examiner insight: in a multi-part question, later parts reuse earlier results, so write each result clearly. Leave the surd exact when the question asks for an exact answer.

3. Midpoint = (-1, 3) [1]. Gradient AB = (5 - 1)/(2 + 4) = 2/3 [1]. Perpendicular gradient = -3/2 [1]. y - 3 = -(3/2)(x + 1), so 2y - 6 = -3x - 3 and 3x + 2y - 3 = 0 [1] [4] Examiner insight: the usual slip is using the gradient of AB itself, or the reciprocal without the sign change. Check that the midpoint satisfies your final equation.

4. (a) Gradient = (31.4 - 14.6)/(6 - 2) = 4.2 [1]. V - 14.6 = 4.2(t - 2) [1]. V = 6.2 + 4.2t [1] (b) The pool gains 4.2 cubic metres each hour [1]. 6.2 is the volume at t = 0, the initial volume [1] (c) 6.2 + 4.2t = 50 gives t = 43.8/4.2 = 73/7 = 10.4 hours [1] (d) For example, the flow may not stay constant, or the model gives volumes above the pool’s capacity if it keeps running [1] [7] Examiner insight: interpretations score best when they use the units and the context (“cubic metres per hour”, “initial volume”), not only the words “gradient” and “intercept”. A limitation should be specific to this situation.

5. (a) r² = (3 + 2)² + (8 - 5)² = 25 + 9 = 34 [1]. (x + 2)² + (y - 5)² = 34 [1] (b) For (0, 10): 4 + 25 = 29 < 34, so inside [1]. For (3, 0): 25 + 25 = 50 > 34, so outside [1] [4] Examiner insight: compare with r² = 34, not with r. Showing the value you calculate for each point makes the conclusion checkable.

6. (a) Divide by 3: x² + y² + 4x - 6y - 1 = 0 [1]. (x + 2)² - 4 + (y - 3)² - 9 = 1 [1]. Centre (-2, 3), radius √14 [1] (b) (x + 6)² + (y - 2)² = 40 - k [1]. This needs 40 - k > 0, so k < 40 [1] (c) 40 - k = 25, so k = 15 [1] [6] Examiner insight: dividing by the common coefficient first is the step that is most often missed. State the condition r² > 0 rather than only quoting the final inequality.

7. (a) C1: (x - 1)² + (y - 2)² = 9, centre (1, 2), radius 3 [1]. C2: (x - 9)² + (y - 2)² = 16, centre (9, 2), radius 4 [1] (b) Distance between centres = 8, and 3 + 4 = 7 < 8, so the circles are separate [1]. Shortest distance = 8 - 7 = 1 [1] [4] Examiner insight: the argument is a comparison of the distance between centres with the sum of the radii, so show both numbers. The smallest gap is measured along the line of centres.

8. (a) Gradient PA = (6 + 1)/(2 - 1) = 7 [1]. Gradient PB = (6 - 5)/(2 - 9) = -1/7 [1]. Product = -1, so angle APB = 90°, and P lies on the circle with diameter AB [1] (b) Centre = (5, 2) [1]. r² = (9 - 5)² + (5 - 2)² = 25, so (x - 5)² + (y - 2)² = 25 [1] [5] Examiner insight: the question bans the equation in part (a), so the argument must come from gradients and the angle in a semicircle. End with a sentence linking a right angle to being on the circle.

9. (a) y = 7 - x, so y + 1 = 8 - x [1]. (x - 2)² + (8 - x)² = 26 gives x² - 10x + 21 = 0 [1]. x = 3 or 7, so A(3, 4) and B(7, 0) [1] (b) Midpoint of AB = (5, 2) [1]. Gradient from the centre (2, -1) to (5, 2) is 1, AB has gradient -1, and 1 × (-1) = -1, so the line from the centre is perpendicular to AB and meets it at its midpoint [1] (c) AB = √(16 + 16) = 4√2 [1] [6] Examiner insight: a gradient check shows the perpendicularity, so give both gradients. The property in part (b) also gives the chord length from the centre’s distance, which is a useful cross-check.

10. (a) (x - 1)² - 1 + (y - 3)² - 9 = 10 [1]. Centre (1, 3), radius √20 = 2√5 [1] (b) 16 + 4 = 20, so (5, 5) is on the circle [1]. Radius gradient = (5 - 3)/(5 - 1) = 1/2, so tangent gradient = -2 [1]. y - 5 = -2(x - 5), giving 2x + y - 15 = 0 [1] (c) X is (15/2, 0) and Y is (0, 15) [1]. Area = (1/2) × (15/2) × 15 = 225/4 [1] [7] Examiner insight: checking that the point lies on the circle is a cheap line of working. The tangent must use the point of contact, not the centre, in the line equation.

11. (a) Substitute: (x - 3)² + (3x + k - 1)² = 10 [1]. Expanding: x² - 6x + 9 + 9x² + 6(k - 1)x + (k - 1)² = 10, so 10x² + (6k - 12)x + k² - 2k = 0 [1] (b) Discriminant = (6k - 12)² - 40(k² - 2k) [1] = 36k² - 144k + 144 - 40k² + 80k = -4k² - 64k + 144 = -4(k - 2)(k + 18) [1] (c) Tangent when the discriminant is 0: k = 2 or k = -18 [1]. Two points when (k - 2)(k + 18) < 0, so -18 < k < 2 [1] (d) With k = 2 the quadratic is 10x² = 0, so x = 0 and y = 2: the point is (0, 2) [1] [7] Examiner insight: in “show that” parts every algebraic step should be on the page, because the result is given. For the range, sketching the sign of the quadratic in k avoids reversing the inequality.

Where marks are usually lost

• Using the negative of the gradient, or its reciprocal, as the perpendicular gradient.
• Stopping at y = mx + c when the question asks for ax + by + c = 0 with integer coefficients.
• Reading the centre of (x + 2)² + (y - 5)² = 34 with the wrong signs.
• Comparing a point’s distance-squared with r instead of r².
• Forgetting to divide by the coefficient of x² and y² before completing the square.
• Taking the circle’s centre rather than the point of contact as the point on the tangent.
• Giving a model’s gradient with no units or no link to the context.
• Rounding too early, or giving a decimal where an exact surd was requested.
• Writing a discriminant condition with the wrong inequality direction when finding a range.

Next steps

• Review the methods in the revision notes.
• Re-read the study guide for any topic that cost you marks.
• See the OCR A Level Mathematics hub and the checklist.
• Take one of our free 10-minute diagnostics to find your next topic.
• Book a free trial class.

Official syllabus

Cambridge OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by Cambridge OCR: section 1.03, Coordinate Geometry in the x-y plane, references 1.03a to 1.03f.