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TMUA Coordinate Geometry: Lines and Circles

Coordinate geometry describes lines and circles with equations. The specification covers gradients, completing the square on a circle and tangents.

Paper 1 and Paper 2 topic. Specification codes: MM3.1, MM3.2, M4.10. Last checked 12 November 2026.

Straight lines

Parallel lines have equal gradients. Perpendicular gradients multiply to −1-1. Use y−y1=m(x−x1)y-y_1=m(x-x_1) to build an equation from a point and a gradient.

Circles

(x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2 has centre (a,b)(a,b) and radius rr. If the equation is x2+y2+cx+dy+e=0x^2+y^2+cx+dy+e=0, complete the square in xx and in yy to read off the centre and radius.

Tangents

A tangent to a circle is perpendicular to the radius at the point of contact, so its gradient is the negative reciprocal of the radius gradient.

Useful facts

The distance between (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) is (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, and the midpoint averages the coordinates. A perpendicular from the centre of a circle to a chord bisects the chord.

To test whether a point is inside, on or outside a circle, compare its squared distance from the centre with r2r^2.

Worked examples

Example 1

Find the equation of the line through (1,2)(1,2) perpendicular to 3x+2y=63x+2y=6.

Show the worked solution

The given line has gradient −32-\tfrac32, so the perpendicular has gradient 23\tfrac23. Then y−2=23(x−1)y-2=\tfrac23(x-1), i.e. 2x−3y+4=02x-3y+4=0.

Answer: 2x−3y+4=02x-3y+4=0

Example 2

Find the centre and radius of x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0.

Show the worked solution

Complete the square: (x−3)2+(y+2)2=12+9+4=25(x-3)^2+(y+2)^2=12+9+4=25. Centre (3,−2)(3,-2), radius 5.

Answer: Centre (3,−2)(3,-2), radius 55

Example 3

Find the tangent to (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25 at the point (4,2)(4,2).

Show the worked solution

The centre is (1,−2)(1,-2), so the radius has gradient 43\tfrac43 and the tangent has gradient −34-\tfrac34. Then y−2=−34(x−4)y-2=-\tfrac34(x-4), so 3x+4y=203x+4y=20.

Answer: 3x+4y=203x+4y=20

Common mistakes

  • Using the reciprocal of a gradient without changing the sign.
  • Reading the centre as (−a,−b)(-a,-b) from (x−a)2+(y−b)2(x-a)^2+(y-b)^2: the signs are the opposite of what is inside the brackets.
  • Forgetting that the right-hand side of the circle equation is r2r^2, not rr.
  • Not rearranging the line to y=mx+cy=mx+c before reading its gradient.

Test yourself on this topic with our free practice questions.

Quick answers

What is the equation of a circle?

(x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2, with centre (a,b)(a,b) and radius rr.

How do you find the gradient of a perpendicular line?

Take the negative reciprocal. If m=23m=\tfrac23 the perpendicular gradient is −32-\tfrac32.

How do you find the equation of a tangent to a circle?

Find the gradient of the radius to the point, take the negative reciprocal, then use the point-gradient form.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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