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Learn9709Paper 1 · Pure Mathematics 1Coordinate geometryThe equation of a circle
Coordinate geometry · Subtopic 3 of 4

The Equation of a Circle

Why (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2 is a circle, how to find the centre and radius from the expanded form by completing the square, and how to use diameters and right angles to find a circle's equation.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 7 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Write down the equation of a circle from its centre and radius, and read them back from the equation.
  • Complete the square to find the centre and radius from x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0.
  • Decide whether a point lies inside, on or outside a circle.
  • Use a diameter, and the angle in a semicircle, to find a circle's equation.
Pause and think

Which points are exactly 55 units from the origin? Write a condition on xx and yy.

Section 2 of 14

Core idea

Start hereEvery point exactly 5 km from the mast

A radio mast reaches exactly 55 km. Which houses are on the edge of its range? Every point at distance 55 from the mast, which is a circle. Using the distance formula, "distance from the centre is 55" becomes an equation, and that equation is the circle.

The big idea

A circle with centre (a,b)(a,b) and radius rr has equation (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2. It's just Pythagoras: the distance from (x,y)(x,y) to the centre equals rr.

1. Deriving the equation

Why the circle equation is Pythagoras
−4−22468−22460xyr = 3centre (2, 2)(x, y)
Wherever (x,y)(x,y) is on the circle, the dashed sides are x−2x-2 and y−2y-2 and the slanted side is 33. So (x−2)2+(y−2)2=9(x-2)^2+(y-2)^2=9.
Circle equation
  1. distance from (x,y) to (a,b)=r\text{distance from }(x,y)\text{ to }(a,b)=r
    Definition of a circle.
  2. (x−a)2+(y−b)2=r\sqrt{(x-a)^2+(y-b)^2}=r
    Distance formula.
  3. (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2
    Square both sides (both are non-negative, so no new solutions appear).
Centre (a,b)(a,b), radius rr. Note the minus signs: (x+1)2(x+1)^2 means a=−1a=-1.
Worked exampleReading and writing circle equations

(a) State the centre and radius of (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25. (b) Write the circle with centre (−1,4)(-1,4), radius 33. (c) Is (6,2)(6,2) inside, on or outside the circle in (a)?

Goal
Centre/radius, an equation, and a position.
Strategy
Compare with (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2; for (c) compare the distance² with r2r^2.
  1. (x−3)2+(y−(−2))2=52(x-3)^2+(y-(-2))^2=5^2
    Rewrite to match the form.

  2. (a) Centre (3,−2)(3,-2), radius 55.

  3. (x+1)2+(y−4)2=9(x+1)^2+(y-4)^2=9
    (b) a=−1a=-1, b=4b=4, r2=9r^2=9.
  4. (6−3)2+(2+2)2=9+16=25(6-3)^2+(2+2)^2=9+16=25
    (c) Equals r2r^2.
Answer(a) (3,−2)(3,-2), 55; (b) (x+1)2+(y−4)2=9(x+1)^2+(y-4)^2=9; (c) on the circle.
✓ Check(6,2)(6,2) is 33 right and 44 up from the centre: a 3-4-5 triangle ✓.

2. Expanded form: complete the square twice

Worked examplex2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0

Find the centre and radius.

Goal
Rewrite as (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2.
Strategy
Group xx terms and yy terms; complete the square on each.
  1. (x2−6x)+(y2+4y)=12(x^2-6x)+(y^2+4y)=12
    Group; move the constant across.
  2. (x−3)2−9+(y+2)2−4=12(x-3)^2-9+(y+2)^2-4=12
    Complete each square.
  3. (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25
    Add 1313.
AnswerCentre (3,−2)(3,-2), radius 55.
✓ CheckExpand back: x2−6x+9+y2+4y+4=25⇒x2+y2−6x+4y−12=0x^2-6x+9+y^2+4y+4=25\Rightarrow x^2+y^2-6x+4y-12=0 ✓.
Shortcutx2+y2+2gx+2fy+c=0:centre (−g,−f), r=g2+f2−cx^2+y^2+2gx+2fy+c=0:\quad\text{centre }(-g,-f),\ r=\sqrt{g^2+f^2-c}
where
g2+f2−c≤0g^2+f^2-c\le0
no real circle (radius would be zero or imaginary).
Edge caseNot a circle The x2x^2 and y2y^2 coefficients must be equal and there must be no xyxy term. x2+2y2=4x^2+2y^2=4 is an ellipse.

3. Finding a circle from geometry

  • The perpendicular bisector of any chord passes through the centre.
  • The angle in a semicircle is 90∘90^\circ, so a right-angled triangle's hypotenuse is a diameter.
  • A tangent is perpendicular to the radius at the contact point.
Worked exampleCircle through (0,0)(0,0), (6,0)(6,0), (0,8)(0,8)

Find the equation of the circle through these three points.

Goal
Centre and radius.
Strategy
Perpendicular bisectors of two chords meet at the centre.

  1. Chord (0,0)(0,0)–(6,0)(6,0) is horizontal; its perpendicular bisector is x=3x=3.

  2. Chord (0,0)(0,0)–(0,8)(0,8) is vertical; its perpendicular bisector is y=4y=4.

  3. centre (3,4),r=32+42=5\text{centre }(3,4),\quad r=\sqrt{3^2+4^2}=5
    Distance to the origin.
Answer(x−3)2+(y−4)2=25(x-3)^2+(y-4)^2=25.
✓ Check(6,0)(6,0): 9+16=259+16=25 ✓; (0,8)(0,8): 9+16=259+16=25 ✓.
Pause and think

Find the centre and radius of x2+y2+8x−2y+8=0x^2+y^2+8x-2y+8=0.

Summary

  • (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2 from Pythagoras.
  • Expanded form: complete both squares.
  • Centre from perpendicular bisectors or a diameter's midpoint.
Key words
Radius
Distance from centre to circle.
Chord
A segment joining two points on the circle.
The circle (x − 2)² + (y − 1)² = 9
−2246−3−2−1123450xyupper halflower halfr = 3centre (2, 1)(−1, 1)
A circle with centre (a,b)(a,b) and radius rr has equation (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2. Every point on it is exactly rr from the centre.
HistoryTwo thousand years of circle-chasing. Greek mathematicians studied circles with ruler and compasses for centuries, but a circle could only be written as an equation after Descartes and Fermat introduced coordinates in the 1630s. Suddenly, questions about where a line meets a circle became questions about solving equations.
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