Home / A-Level Maths 9709 / Coordinate geometry / The equation of a circle
Cambridge International AS & A Level · Topic 1.3
A-Level Maths 9709 / Topic 1.3 / AS & A Level

Pythagoras, all the way round.

Every point on a circle is the same distance from its centre, and Pythagoras turns that fact into an equation. Read the centre and radius from either form of the equation, then use circle geometry to find tangents and circles through given points.

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The key idea

\((x - a)^{2} + (y - b)^{2} = r^{2}\) is the circle with centre \((a, b)\) and radius \(r\). For \(x^{2} + y^{2} + 2gx + 2fy + c = 0\), complete the square: the centre is \((-g, -f)\) and the radius is \(\sqrt{g^{2} + f^{2} - c}\).

x y C(3, −2) P(6, 2) r = 5 (x − 3)² + (y + 2)² = 25 radius CP: gradient ⁴⁄₃ tangent at P: gradient −³⁄₄ 3x + 4y − 26 = 0
Fig. 1: the circle with centre C(3, -2) and radius 5 passes through P(6, 2); the tangent at P is perpendicular to the radius CP
Section 01

Centre and radius.

A point \((x, y)\) is on the circle when its distance from the centre \((a, b)\) is \(r\). By Pythagoras that distance is \(\sqrt{(x - a)^{2} + (y - b)^{2}}\); squaring gives the equation of the circle.

Worked example Write the equation
Find the equation of the circle with centre \((3, -2)\) and radius 5, and show that \(P(6, 2)\) lies on it (Fig. 1).
  1. \(a = 3,\ b = -2,\ r = 5\): \((x - 3)^{2} + (y - (-2))^{2} = 5^{2}\).
  2. So \((x - 3)^{2} + (y + 2)^{2} = 25\).
  3. At \(P\): \((6 - 3)^{2} + (2 + 2)^{2} = 9 + 16 = 25\), so \(P\) is on the circle.
Answer: \((x - 3)^{2} + (y + 2)^{2} = 25\); \(P\) lies on it
Worked example From a diameter
\(A(-1, 2)\) and \(B(7, 8)\) are the ends of a diameter. Find the equation of the circle.
  1. The centre is the midpoint of the diameter: \(\left(\dfrac{-1 + 7}{2},\ \dfrac{2 + 8}{2}\right) = (3, 5)\).
  2. \(AB = \sqrt{8^{2} + 6^{2}} = 10\), so the radius is 5.
  3. Substitute into \((x - a)^{2} + (y - b)^{2} = r^{2}\).
Answer: \((x - 3)^{2} + (y - 5)^{2} = 25\)
Your turn
Find the equation of the circle with centre \((2, -1)\) that passes through \((5, 3)\).
Examiner trap

The right-hand side is \(r^{2}\), not \(r\): a circle of radius 5 has \(= 25\), and \((x - 3)^{2} + (y + 2)^{2} = 12\) has radius \(\sqrt{12} = 2\sqrt{3}\), not 12.

Section 02

The expanded form.

Expanding the brackets gives \(x^{2} + y^{2} + 2gx + 2fy + c = 0\). To read it, complete the square in \(x\) and in \(y\) separately, then move the constants to the right.

Worked example Complete two squares
Find the centre and radius of \(x^{2} + y^{2} - 6x + 4y - 12 = 0\).
  1. Group: \((x^{2} - 6x) + (y^{2} + 4y) = 12\).
  2. Complete each square: \((x - 3)^{2} - 9 + (y + 2)^{2} - 4 = 12\).
  3. \((x - 3)^{2} + (y + 2)^{2} = 25\).
Answer: Centre \((3, -2)\), radius 5 (the circle in Fig. 1)
Worked example Using the formula
Find the centre and radius of \(x^{2} + y^{2} + 8x - 2y + 8 = 0\).
  1. Compare with \(x^{2} + y^{2} + 2gx + 2fy + c = 0\): \(2g = 8\), \(2f = -2\), \(c = 8\), so \(g = 4\), \(f = -1\).
  2. Centre \((-g, -f) = (-4, 1)\).
  3. Radius \(\sqrt{g^{2} + f^{2} - c} = \sqrt{16 + 1 - 8} = \sqrt{9}\).
Answer: Centre \((-4, 1)\), radius 3
Worked example Not every equation is a circle
Explain why \(x^{2} + y^{2} - 2x + 4y + 10 = 0\) does not represent a circle.
  1. \((x - 1)^{2} + (y + 2)^{2} = -10 + 1 + 4 = -5\).
  2. A sum of squares cannot be negative, so no point satisfies the equation.
Answer: \(r^{2} = -5 < 0\): there is no such circle
Examiner trap

The coefficients of \(x^{2}\) and \(y^{2}\) must be 1 before you read off \(g\) and \(f\). For \(2x^{2} + 2y^{2} - 8x + 12y - 6 = 0\), divide by 2 first. And watch the signs: \(-6x\) gives centre \(x = +3\).

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Find the centre and radius of \((x + 1)^{2} + (y - 4)^{2} = 9\).

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Which is the equation of the circle with centre \((2, -5)\) and radius 4?

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Find the centre of \(x^{2} + y^{2} - 10x + 6y + 9 = 0\).

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\(P\) is a point on a circle with centre \(C\). The tangent at \(P\) is perpendicular to

Section 03

Tangents and circle geometry.

In the lab, move the point of contact around the circle and check that the tangent always meets the radius at \(90^{\circ}\). Three facts from circle geometry do most of the work in 9709 questions.

Worked example Tangent at a point
Find the equation of the tangent to \((x - 3)^{2} + (y + 2)^{2} = 25\) at \(P(6, 2)\) (Fig. 1).
  1. Gradient of radius \(CP = \dfrac{2 - (-2)}{6 - 3} = \dfrac{4}{3}\).
  2. The tangent is perpendicular to \(CP\): gradient \(-\dfrac{3}{4}\).
  3. \(y - 2 = -\dfrac{3}{4}(x - 6) \Rightarrow 4y - 8 = -3x + 18\).
Answer: \(3x + 4y - 26 = 0\)
Worked example Circle through three points
Find the equation of the circle through \(A(1, 1)\), \(B(7, 1)\) and \(C(1, 5)\).
  1. \(AB\) is horizontal and \(AC\) is vertical, so angle \(BAC = 90^{\circ}\).
  2. By the angle in a semicircle, \(BC\) is a diameter. Its midpoint is the centre: \((4, 3)\).
  3. \(r^{2} = (7 - 4)^{2} + (1 - 3)^{2} = 9 + 4 = 13\).
Answer: \((x - 4)^{2} + (y - 3)^{2} = 13\)
Examiner trap

The tangent's gradient is the negative reciprocal of the RADIUS gradient, not of the gradient from the origin to \(P\). Always use the centre, which is only the origin for \(x^{2} + y^{2} = r^{2}\).

Stage 2 · Exam

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Find the centre and radius of the circle \(x^{2} + y^{2} + 6x - 8y - 11 = 0\).

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\(A(-2, 3)\) and \(B(6, -3)\) are the ends of a diameter of a circle. Find its equation.

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Find the equation of the tangent to \((x - 1)^{2} + (y - 3)^{2} = 25\) at the point \((5, 6)\).

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A circle has its centre on the \(x\)-axis and passes through \((1, 4)\) and \((5, 2)\). Find its equation.

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Circles written and read.

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Go deeper · practice
Six original Cambridge-style questions
Writing and expanding a circle's equation, a centre and radius after dividing through, a tangent, a circle from the angle in a semicircle, a point outside a circle, and the area watered by a sprinkler.