AS & A Level · Practice questions · The equation of a circle

Centre, radius, right angles.

Six original Cambridge-style questions on the equation of a circle: both forms, tangents, circles through three points and a context problem with a garden sprinkler.

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

The method

Get to \((x - a)^{2} + (y - b)^{2} = r^{2}\).

Centre and radius give the equation directly; remember \(r^{2}\) on the right.
Expanded form: make the \(x^{2}\) and \(y^{2}\) coefficients 1, then complete both squares.
Diameter: centre = midpoint, radius = half the length.
Tangent at \(P\): perpendicular to the radius \(CP\).
Inside or outside: compare the distance from the centre with \(r\).

Recall

[2 marks]

Write down the equation of the circle with centre \((-3, 1)\) and radius \(\sqrt{7}\), and express it in the form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).

\((x + 3)^{2} + (y - 1)^{2} = 7\). ✓
\(x^{2} + 6x + 9 + y^{2} - 2y + 1 = 7 \Rightarrow x^{2} + y^{2} + 6x - 2y + 3 = 0\). ✓

Calculation

[3 marks]

Find the centre and radius of the circle \(2x^{2} + 2y^{2} - 8x + 12y - 6 = 0\).

Divide by 2: \(x^{2} + y^{2} - 4x + 6y - 3 = 0\). ✓
\((x - 2)^{2} - 4 + (y + 3)^{2} - 9 = 3 \Rightarrow (x - 2)^{2} + (y + 3)^{2} = 16\). ✓
Centre \((2, -3)\), radius 4. ✓

Calculation

[4 marks]

The circle \(x^{2} + y^{2} - 4x - 2y - 20 = 0\) passes through \(P(-1, 5)\). Find the equation of the tangent to the circle at \(P\).

\((x - 2)^{2} + (y - 1)^{2} = 25\): centre \(C(2, 1)\). ✓
Gradient \(CP = \dfrac{5 - 1}{-1 - 2} = -\dfrac{4}{3}\), so the tangent gradient is \(\dfrac{3}{4}\). ✓
\(y - 5 = \dfrac{3}{4}(x + 1) \Rightarrow 4y - 20 = 3x + 3\). ✓
\(3x - 4y + 23 = 0\). ✓

Analysis

[5 marks]

The points \(A(-1, 1)\), \(B(9, 1)\) and \(C(1, 5)\) are given.

(a) Show that angle \(ACB = 90^{\circ}\).

(b) Hence find the equation of the circle through \(A\), \(B\) and \(C\).

(a) Gradient \(AC = \dfrac{4}{2} = 2\), gradient \(BC = \dfrac{4}{-8} = -\dfrac{1}{2}\); \(2 \times \left(-\dfrac{1}{2}\right) = -1\), so \(AC \perp BC\). ✓
(b) By the angle in a semicircle, \(AB\) is a diameter: centre \((4, 1)\), radius \(\dfrac{10}{2} = 5\). ✓
\((x - 4)^{2} + (y - 1)^{2} = 25\) (check \(C\): \(9 + 16 = 25\)). ✓
(c) \(x = 0\): \(16 + (y - 1)^{2} = 25 \Rightarrow (y - 1)^{2} = 9 \Rightarrow y = 4\) or \(y = -2\). ✓
The circle crosses the \(y\)-axis at \((0, 4)\) and \((0, -2)\). ✓

Proof

[3 marks]

Show that the point \((7, 3)\) lies outside the circle \(x^{2} + y^{2} - 6x + 2y - 15 = 0\).

\((x - 3)^{2} + (y + 1)^{2} = 15 + 9 + 1 = 25\): centre \((3, -1)\), radius 5. ✓
Distance from the centre to \((7, 3)\): \(\sqrt{4^{2} + 4^{2}} = \sqrt{32}\). ✓
\(\sqrt{32} > \sqrt{25} = 5\), so the point is outside the circle. ✓

Analysis

[5 marks]

Coordinates are in metres. A sprinkler at \(S(4, 3)\) waters all points within 5 m of it. A straight path runs along the \(x\)-axis.

(a) Write down the equation of the edge of the watered region.

(b) Find the length of path that gets watered.

(d) The sprinkler is moved along the line \(x = 4\), away from the path, until the watered circle just touches the path. State the new equation of the edge.

(a) \((x - 4)^{2} + (y - 3)^{2} = 25\). ✓
(b) \(y = 0\): \((x - 4)^{2} = 16 \Rightarrow x = 0\) or \(x = 8\); the watered length is 8 m. ✓
(c) Distance\(^{2}\) from \(S\) to the tree \(= 5^{2} + 3^{2} = 34 > 25\), so the tree is outside the circle and is not watered. ✓
(d) The path is a tangent, so the centre is 5 m above it: \(S(4, 5)\). ✓
New edge: \((x - 4)^{2} + (y - 5)^{2} = 25\). ✓

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.