IGCSE · Practice questions · The equation of a circle

Find the centre, find the radius.

Six original Cambridge-style questions on the equation of a circle: centre-radius and general forms, completing the square, diameters, points on the circle, the circle through three points and the condition for a real circle, all without a calculator.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Centre and \(r^{2}\), then the equation.

Equation: \((x - a)^{2} + (y - b)^{2} = r^{2}\); the bracket signs are opposite to the centre's coordinates.
Radius: from a point, \(r^{2}\) is the distance squared from the centre; from a diameter, the centre is the midpoint.
General form: divide so \(x^{2}\) and \(y^{2}\) have coefficient 1, then complete the square or use centre \((-g, -f)\), \(r = \sqrt{g^{2} + f^{2} - c}\).
Real circle: \(g^{2} + f^{2} - c > 0\).
Three points: two perpendicular bisectors of chords meet at the centre; a right angle at a point on the circle means the opposite side is a diameter.

Recall

[2 marks]

Write down the centre and radius of each circle.

(a) \((x - 6)^{2} + (y + 1)^{2} = 81\) (b) \(x^{2} + (y + 4)^{2} = 12\)

(a) Centre \((6, -1)\), radius \(\sqrt{81} = 9\). ✓
(b) Centre \((0, -4)\), radius \(\sqrt{12} = 2\sqrt{3}\). ✓

Calculation

[3 marks]

A circle has centre \((-1, 4)\) and passes through the point \((3, 1)\). Find the equation of the circle, giving your answer in the form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).

\(r^{2} = (3 - (-1))^{2} + (1 - 4)^{2} = 16 + 9 = 25\). ✓
\((x + 1)^{2} + (y - 4)^{2} = 25\). ✓
Expand: \(x^{2} + 2x + 1 + y^{2} - 8y + 16 - 25 = 0\), so \(x^{2} + y^{2} + 2x - 8y - 8 = 0\). ✓

Calculation

[4 marks]

A circle has equation \(x^{2} + y^{2} - 10x + 4y + 4 = 0\).

(a) By completing the square, find the centre and the radius of the circle.

(b) Show that the circle touches the \(y\)-axis, and state the point where it does so.

(a) \((x - 5)^{2} - 25 + (y + 2)^{2} - 4 + 4 = 0\), so \((x - 5)^{2} + (y + 2)^{2} = 25\). ✓
Centre \((5, -2)\), radius 5. ✓
(b) The distance from the centre to the \(y\)-axis is the \(x\)-coordinate of the centre, 5, which equals the radius, so the circle just touches the \(y\)-axis. ✓
Check: put \(x = 0\): \(y^{2} + 4y + 4 = 0\), so \((y + 2)^{2} = 0\) and \(y = -2\) only. The circle touches the \(y\)-axis at \((0, -2)\). ✓

Calculation

[5 marks]

The points \(A(-3, -1)\) and \(B(5, 5)\) are the ends of a diameter of a circle.

(a) Find the equation of the circle.

(b) Show that the point \(C(4, 6)\) lies on the circle.

(a) Centre = midpoint of \(AB\) = \(\left(\dfrac{-3 + 5}{2}, \dfrac{-1 + 5}{2}\right) = (1, 2)\). ✓
\(AB = \sqrt{8^{2} + 6^{2}} = 10\), so \(r = 5\): \((x - 1)^{2} + (y - 2)^{2} = 25\). ✓
(b) \((4 - 1)^{2} + (6 - 2)^{2} = 9 + 16 = 25\), so \(C\) is on the circle. ✓
(c) Gradient of \(AC = \dfrac{6 - (-1)}{4 - (-3)} = 1\); gradient of \(BC = \dfrac{6 - 5}{4 - 5} = -1\). The product is \(-1\), so \(AC\) is perpendicular to \(BC\). ✓
This was expected: \(AB\) is a diameter and \(C\) is on the circle, so angle \(ACB\) is the angle in a semicircle, \(90^{\circ}\). ✓

Analysis

[5 marks]

A circle passes through the points \(A(-1, 1)\), \(B(7, 1)\) and \(C(3, -7)\).

(a) Find the equation of the perpendicular bisector of \(AC\).

(b) Hence find the centre and radius of the circle, and its equation.

(a) Midpoint of \(AC\) = \((1, -3)\); gradient of \(AC = \dfrac{-7 - 1}{3 - (-1)} = -2\), so the perpendicular gradient is \(\dfrac{1}{2}\). ✓
Perpendicular bisector: \(y + 3 = \dfrac{1}{2}(x - 1)\), that is \(y = \dfrac{1}{2}x - \dfrac{7}{2}\). ✓
(b) \(AB\) is horizontal with midpoint \((3, 1)\), so its perpendicular bisector is \(x = 3\). ✓
Put \(x = 3\) into (a): \(y = \dfrac{3}{2} - \dfrac{7}{2} = -2\). The centre is \((3, -2)\). ✓
\(r^{2} = (-1 - 3)^{2} + (1 + 2)^{2} = 16 + 9 = 25\), so \(r = 5\) and the equation is \((x - 3)^{2} + (y + 2)^{2} = 25\). Check with \(C\): \(0 + (-5)^{2} = 25\). ✓

Analysis

[6 marks]

The equation \(x^{2} + y^{2} - 2kx + 6y + 2k + 1 = 0\), where \(k\) is a constant, represents a curve \(S\).

(a) Find, in terms of \(k\), the coordinates of the centre and an expression for \(r^{2}\).

(b) Show that \(S\) is a circle for every value of \(k\).

(d) Find the value of \(k\) for which \(S\) passes through the origin.

(a) Complete the squares: \((x - k)^{2} - k^{2} + (y + 3)^{2} - 9 + 2k + 1 = 0\). ✓
So \((x - k)^{2} + (y + 3)^{2} = k^{2} - 2k + 8\): centre \((k, -3)\) and \(r^{2} = k^{2} - 2k + 8\). ✓
(b) \(k^{2} - 2k + 8 = (k - 1)^{2} + 7\). Since \((k - 1)^{2} \ge 0\), \(r^{2} \ge 7 > 0\), so \(S\) is always a real circle. ✓
(c) The least value of \(r^{2}\) is 7, when \(k = 1\), so the least radius is \(\sqrt{7}\). ✓
(d) Put \(x = 0\), \(y = 0\): \(2k + 1 = 0\), so \(k = -\dfrac{1}{2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.