A Level Maths · Topic guide

Pure: Coordinate Geometry (Lines and Circles)

In A-level Pure Mathematics, coordinate geometry of lines and circles is the study of straight lines and circles using algebra on the x-y plane, covering gradients, midpoints, distances, equations of lines and circles, and conditions for tangency or intersection. Circle questions often use the fact that a tangent is perpendicular to the radius at the point of contact.

A LevelPureEdexcelAQAOCRWJEC

Before you start

Make sure you're comfortable with these topics first:

Method

  1. Use the gradient formula (y2-y1)/(x2-x1) to find gradients between two points; parallel lines share a gradient, and perpendicular lines have gradients that multiply to give -1.
  2. Use the midpoint formula ((x1+x2)/2, (y1+y2)/2) and the distance formula sqrt((x2-x1)^2+(y2-y1)^2) whenever a question mentions the middle or the length of a line segment.
  3. Write a line's equation using y - y1 = m(x - x1), then rearrange into the requested form (y = mx + c or ax + by + c = 0).
  4. Recognise a circle equation in the form (x - a)^2 + (y - b)^2 = r^2 with centre (a, b) and radius r; complete the square if the equation is given in expanded form.
  5. For tangent problems, find the gradient of the radius to the point of contact first, then take the negative reciprocal to get the tangent gradient.
  6. For intersection or tangency between a line and a circle, substitute the line into the circle equation and use the discriminant: two solutions means the line is a secant, one means it is a tangent, and none means it misses the circle.

Worked example

A circle has centre (3, 2) and passes through the point (7, 5). Find the equation of the circle, and determine whether the point (0, -1) lies inside, on, or outside the circle.

  1. Find the radius using the distance formula: r = sqrt((7-3)^2 + (5-2)^2) = sqrt(16 + 9) = sqrt(25) = 5.
  2. Write the equation of the circle: (x - 3)^2 + (y - 2)^2 = 25.
  3. Substitute (0, -1) into the left-hand side: (0-3)^2 + (-1-2)^2 = 9 + 9 = 18.
  4. Compare 18 with r^2 = 25: since 18 < 25, the point is closer to the centre than the radius allows.
  5. Final answer: the circle has equation (x - 3)^2 + (y - 2)^2 = 25, and the point (0, -1) lies inside the circle.

Practice questions

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Q1Points A(2, 3) and B(6, 11) are joined by a line. Find the gradient of AB.Show answer

Answer: 2 (gradient = (11-3)/(6-2)).

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Q2Find the midpoint of C(-4, 5) and D(2, -3).Show answer

Answer: (-1, 1).

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Q3Find the equation of the line through (1, 4) with gradient -3, in the form y = mx + c.Show answer

Answer: y = -3x + 7.

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Q4A circle has centre (1, -2) and radius 6. Write down its equation.Show answer

Answer: (x - 1)^2 + (y + 2)^2 = 36.

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Q5Find the equation of the perpendicular bisector of the segment joining E(0, 6) and F(8, 2), in the form ax + by + c = 0.Show answer

Answer: 2x - y - 4 = 0 (midpoint (4,4), perpendicular gradient 2).

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Q6A circle has equation x^2 + y^2 - 8x + 2y - 8 = 0. Find its centre and radius, and determine whether the point (9, 1) lies inside, on or outside the circle.Show answer

Answer: Centre (4, -1), radius 5; the point (9,1) gives 29 > 25, so it lies outside the circle.

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Exam-style questions

Written in the style of a A Level Maths exam paper, with a full mark scheme.

Q1[4 marks]

A circle has centre (-2, 5) and passes through the point (2, 8). Find the radius of the circle, and write down the equation of the circle.

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Q2[5 marks]

The points P(0, 5) and Q(6, 1) lie on a circle, and the centre of the circle lies on the line y = 2x - 4. Find an equation of the perpendicular bisector of PQ, and hence find the coordinates of the centre of the circle.

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Q3[6 marks]

A circle C has equation (x - 2)^2 + (y + 3)^2 = 20. The point A(6, -1) lies on the circle. Find an equation of the tangent to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

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See real past-paper questions on pure: coordinate geometry (lines and circles), organised by topic with official mark schemes

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