Pure: Coordinate Geometry (Lines and Circles)
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.
Method
- 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.
- 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.
- 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).
- 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.
- 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.
- 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.
- Find the radius using the distance formula: r = sqrt((7-3)^2 + (5-2)^2) = sqrt(16 + 9) = sqrt(25) = 5.
- Write the equation of the circle: (x - 3)^2 + (y - 2)^2 = 25.
- Substitute (0, -1) into the left-hand side: (0-3)^2 + (-1-2)^2 = 9 + 9 = 18.
- Compare 18 with r^2 = 25: since 18 < 25, the point is closer to the centre than the radius allows.
- 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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Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
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.
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.
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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