Equation of a circle and tangents - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Equation of a circle and tangents for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A circle has equation \(x^2 + y^2 = 18\).

Find the radius of the circle. Give your answer in the form \(a\sqrt{2}\), where \(a\) is an integer.

Model answer

\(r = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)

Explanation

The radius is the square root of the number on the right-hand side. Simplify the surd by taking out the largest square factor: 18 = 9 × 2.

Mark scheme

1 mark
M1 for \(\sqrt{18}\)
correct answer with no working scores full marks
1 mark
A1 for \(3\sqrt{2}\)
cao

Question 2

2 marks

Here are four points.
\((5, 5)\), \((7, 1)\), \((-1, -7)\), \((6, 4)\)

Three of the points lie on the circle \(x^2 + y^2 = 50\). Which point does not? You must show your working.

Model answer

\((5, 5)\): \(25 + 25 = 50\) ✓
\((7, 1)\): \(49 + 1 = 50\) ✓
\((-1, -7)\): \(1 + 49 = 50\) ✓
\((6, 4)\): \(36 + 16 = 52\) ✗
The point \((6, 4)\) does not lie on the circle.

Explanation

Substitute each point into \(x^2 + y^2\) and compare with 50. Since 52 is greater than 50, the point \((6, 4)\) is just outside the circle.

Mark scheme

1 mark
M1 for testing at least two points, e.g. \(5^2 + 5^2 = 50\), \(6^2 + 4^2 = 52\)
1 mark
A1 for \((6, 4)\)
with \(36 + 16 = 52\) shown

Question 3

3 marks

The point \((-6, 8)\) lies on the circle \(x^2 + y^2 = 100\).

Find the coordinates of the point where the tangent to the circle at \((-6, 8)\) crosses the y-axis.

Model answer

Gradient of radius \(= \dfrac{8}{-6} = -\dfrac{4}{3}\)
Gradient of tangent \(= \dfrac{3}{4}\)
\(8 = \dfrac{3}{4} \times (-6) + c\)
\(8 = -4.5 + c\), so \(c = 12.5\)
The tangent crosses the y-axis at \((0, 12.5)\).

Explanation

The y-intercept of the tangent is the value of \(c\) in \(y = mx + c\). Find the tangent's gradient from the perpendicular radius, then substitute the point. Take care with signs: \(\frac{3}{4} \times (-6) = -4.5\).

Mark scheme

1 mark
M1 for the gradient of the radius, \(-\frac{4}{3}\)
correct answer with no working scores full marks
1 mark
M1 for the tangent gradient \(\frac{3}{4}\) used with \((-6, 8)\)
1 mark
A1 for \((0, 12.5)\)
oe, e.g. \(\left(0, \frac{25}{2}\right)\)

Question 4

4 marks

Show that the line \(2x + y = 15\) is a tangent to the circle \(x^2 + y^2 = 45\), and find the coordinates of the point where it touches the circle.

Model answer

\(y = 15 - 2x\)
\(x^2 + (15 - 2x)^2 = 45\)
\(x^2 + 225 - 60x + 4x^2 = 45\)
\(5x^2 - 60x + 180 = 0\)
\(x^2 - 12x + 36 = 0\)
\((x - 6)^2 = 0\), so \(x = 6\) only (a repeated root), and \(y = 15 - 12 = 3\).
The line meets the circle at exactly one point, so it is a tangent. It touches the circle at \((6, 3)\).

Explanation

A line can meet a circle at two points, one point (a tangent) or none. Substituting gives a quadratic; a repeated root means exactly one meeting point. You can check with gradients: the radius to \((6, 3)\) has gradient \(\frac{1}{2}\) and the line has gradient \(-2\), which are perpendicular.

Mark scheme

1 mark
M1 for substituting \(y = 15 - 2x\) into the circle, \(x^2 + (15 - 2x)^2 = 45\)
1 mark
M1 for simplifying to \(5x^2 - 60x + 180 = 0\) or \(x^2 - 12x + 36 = 0\)
1 mark
A1 for \((x - 6)^2 = 0\), a repeated root
1 mark
C1 for concluding that the line meets the circle at only one point, so it is a tangent, touching at \((6, 3)\)

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