Ch 3 · 1.3 Coordinate Geometry · A Level Mathematics 9709
Topic 3 of 8 in Pure Mathematics 1 — A Level Mathematics 9709 — 72 lessons.
Introduction
A satellite-navigation app draws your route as straight segments and the range of a phone mast as a circle. Coordinate geometry turns such pictures into equations that you can solve. This lesson covers straight lines and circles, and how solving equations simultaneously finds where graphs meet. (Syllabus section 1.3, Paper 1.)
Key ideas — Gradient, distance and midpoint
For points A(x_1, y_1) and B(x_2, y_2):
🧑🏫 Teacher's explanation
Imagine you are looking at a map of a city laid out on a grid. If you want to walk from one shop to another, the 'gradient' tells you how steep the hill is between them. The 'distance' is the direct, straight-line length of your path, and the 'midpoint' is the exact halfway cafe where you might meet a friend.
📖 Definitions
🔍 Worked example — Easy
Find the gradient and the midpoint of the line segment joining A(2, -3) and B(6, 5).
🔍 Worked example — Standard
Find the equation of the perpendicular bisector of the line segment joining P(1, 2) and Q(5, 10).
🔍 Worked example — Exam-style
A circle has a diameter with endpoints A(-2, 4) and B(4, -2). Find the equation of the circle in the form (x-a)^2 + (y-b)^2 = r^2.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Equations of a straight line
Parallel lines have equal gradients. Perpendicular lines have gradients with product -1: m_1 m_2 = -1, i.e. m_2 = -1/m_1.
🧑🏫 Teacher's explanation
Think of a straight line as a path with a constant 'steepness'. This steepness is the gradient, m. If you know how steep the path is and one single point on it, you can predict every other point on that path. It is like knowing you are walking up a hill at a 30^° angle and you are currently at a height of 10m; you can calculate exactly how high you will be after walking 5m horizontally.
📖 Definitions
🔍 Worked example — Easy
Find the equation of a line passing through the points (2, 5) and (4, 9). Give your answer in the form y = mx + c.
🔍 Worked example — Standard
Find the equation of the line perpendicular to 3x - 2y = 6 that passes through the point (4, 1). Give your answer in the form ax + by + c = 0 where a, b, c are integers.
🔍 Worked example — Exam-style
A line L_1 is parallel to y = 4x - 7 and passes through the intersection of y = 2x + 1 and y = -x + 7. Find the equation of L_1 in the form ax + by + c = 0.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find the gradient of L. [1] (b) Find the equation of the line perpendicular to L that passes through the point (4, 6). Give your answer in the form ax + by + c = 0 where a, b, c are integers. [3] [4 marks]
Key ideas — The equation of a circle
The circle with centre (a, b) and radius r is
🧑🏫 Teacher's explanation
Imagine you are standing in a field at a specific spot, and you are told to walk exactly 5 metres in any direction. If you walk all possible directions and mark your footprints, you will have traced out a circle. In coordinate geometry, the 'spot' is your centre (a, b), and the 'distance' is your radius r. The equation (x-a)^2 + (y-b)^2 = r^2 is simply a way of saying: 'The distance between any point (x, y) on the edge and the centre (a, b) is always r'. This is just the Pythagorean theorem in disguise: a^2 + b^2 = c^2.
📖 Definitions
🔍 Worked example — Easy
Find the equation of a circle with centre (3, -4) and radius 5.
🔍 Worked example — Standard
Find the centre and radius of the circle with equation x^2 + y^2 + 8x - 2y - 8 = 0.
🔍 Worked example — Exam-style
A circle has the equation x^2 + y^2 - 6x + 10y + 9 = 0. Find the coordinates of the centre and the length of the radius.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Circle properties you use
🧑🏫 Teacher's explanation
Imagine a circle as a perfectly round track. If you are standing on the edge and look straight at the center, your line of sight is the radius. If you then turn 90^° and walk along the edge, you are moving along a tangent. This 'perpendicular' relationship is the golden rule of circle geometry.
📖 Definitions
🔍 Worked example — Easy
A circle has the equation (x - 3)^2 + (y + 2)^2 = 25. Find the coordinates of the center and the radius.
🔍 Worked example — Standard
A circle has a diameter with endpoints A(2, 4) and B(8, 12). Find the equation of the circle.
🔍 Worked example — Exam-style
The line y = x + k is a tangent to the circle x^2 + y^2 = 5. Find the possible values of k.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find the equation of the circle in the form (x-a)^2 + (y-b)^2 = r^2. (b) Find the equation of the tangent to the circle at the point (5, 1). [5 marks]
Key ideas — Intersections and the discriminant
The x-coordinates where two graphs meet are the solutions of the equation you get by eliminating y. When this is a quadratic, the discriminant tells you how many meeting points there are:
🧑🏫 Teacher's explanation
Imagine you are driving a car along a straight road (a line) and you want to know if you will ever cross a circular park (a circle). You can describe the road with one equation and the park with another. To find out if you hit the park, you 'merge' the two equations into one single equation that describes the x-coordinates where you are in both places at once.
📖 Definitions
🔍 Worked example — Easy
Find the number of intersection points between the line y = 2x + 3 and the curve y = x^2 + 5x + 1.
🔍 Worked example — Standard
Find the value of k for which the line y = kx - 5 is a tangent to the circle x^2 + y^2 = 25.
🔍 Worked example — Exam-style
The line y = mx + 1 intersects the curve y^2 = 4x at two distinct points. Find the range of values of m.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: midpoint, length and perpendicular bisector
A is (1, 2) and B is (5, 10). Find the length of AB and the equation of the perpendicular bisector of AB.
🔍 Worked example — Easy
Find the length of the line segment joining P(2, 3) and Q(5, 7).
🔍 Worked example — Exam-style
The line segment CD has endpoints C(-3, 4) and D(5, 1). Find the equation of the perpendicular bisector of CD in the form ax + by + c = 0, where a, b, c are integers.
⚠️ Common errors
Worked examples — Example 2: centre, radius and tangent
A circle has equation x^2 + y^2 - 6x + 4y - 12 = 0. Find its centre and radius, and the equation of the tangent at P(6, 2).
🔍 Worked example — Easy
Find the centre and radius of the circle with equation x^2 + y^2 + 8x - 2y - 8 = 0.
🔍 Worked example — Standard
A circle has centre C(3, -2) and radius 5. Find the equation of the tangent to the circle at the point P(6, 2).
🔍 Worked example — Exam-style
The line y = mx + 1 is a tangent to the circle x^2 + y^2 = 5. Find the possible values of m.
⚠️ Common errors
Worked examples — Example 3: circle from a diameter
A(-1, 3) and B(5, -5) are the ends of a diameter. Centre = midpoint = (2, -1); radius = √(3^2 + 4^2) = 5. The circle is (x-2)^2 + (y+1)^2 = 25.
🔍 Worked example — Easy
Find the equation of a circle where the diameter endpoints are (0, 0) and (6, 8).
🔍 Worked example — Standard
The diameter of a circle has endpoints A(-2, 4) and B(4, 2). Find the equation of the circle in the form x^2 + y^2 + 2gx + 2fy + c = 0.
🔍 Worked example — Exam-style
A circle has a diameter with endpoints P(-1, 3) and Q(5, -5). Find the equation of the circle.
⚠️ Common errors
Worked examples — Example 4: when does a line meet a curve?
Find the set of values of k for which the line y = x + k meets the curve y = x^2 - 3x + 5 at two distinct points.
🔍 Worked example — Easy
Find the values of k for which the line y = 2x + k meets the curve y = x^2 at two distinct points.
🔍 Worked example — Standard
The line y = x + k is a tangent to the curve y = x^2 - 4x + 7. Find the value of k.
🔍 Worked example — Exam-style
Find the set of values of k for which the line y = 3x + k does not meet the curve y = x^2 + 2x + 5.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) y = -3x + 5. (2) x + 3(2x - 3) = 12 ⇒ x = 3, y = 3: (3, 3). (3) (x+4)^2 + (y-1)^2 = 25: centre (-4, 1), radius 5. (4) 16 + 9 = 25 ✓; radius gradient tfrac34, tangent gradient -tfrac43: 4x + 3y = 25. (5) x^2 - kx + 4 = 0, k^2 - 16 = 0, so k = ± 4. (6) gradient PR = tfrac15, gradient QR = -5, product -1 so PR ⊥ QR; then PQ is a diameter: centre (3, 2), radius √(13), circle (x-3)^2 + (y-2)^2 = 13.