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

Practise 29 exam-style questions on Equation of a circle and tangents for Edexcel GCSE Maths (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.

29 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

1 mark

Which of these is the equation of a circle with centre (0, 0) and radius 8?

\(x^2 + y^2 = 8\) \(x^2 + y^2 = 16\) \(x^2 + y^2 = 64\) \(x + y = 64\)

Model answer

\(x^2 + y^2 = r^2 = 8^2 = 64\)
The equation is \(x^2 + y^2 = 64\)

Explanation

A circle with centre the origin and radius \(r\) has equation \(x^2 + y^2 = r^2\). With \(r = 8\), the right-hand side is 64. The equation \(x + y = 64\) is a straight line, not a circle.

Mark scheme

1 mark
B1 for \(x^2 + y^2 = 64\)
cao

Question 2

2 marks

Ella is finding the equation of the tangent to the circle \(x^2 + y^2 = 34\) at the point (3, 5).
Here is her working.

Gradient of tangent \(= \dfrac{5}{3}\)
\(y - 5 = \dfrac{5}{3}(x - 3)\)
\(y = \dfrac{5}{3}x\)

What mistake has Ella made?
Write down the correct gradient of the tangent.

Model answer

\(\frac{5}{3}\) is the gradient of the radius from O(0, 0) to (3, 5).
The tangent is perpendicular to the radius, so its gradient is the negative reciprocal.
Gradient of tangent \(= -\dfrac{3}{5}\)
(The tangent is \(y = -\frac{3}{5}x + \frac{34}{5}\).)

Explanation

The gradient from the centre to the point gives the radius, not the tangent. Because a tangent meets the radius at 90°, its gradient is \(-1 \div \frac{5}{3} = -\frac{3}{5}\). A useful check: Ella's line passes through the origin, but a tangent can never pass through the centre of a circle.

Mark scheme

1 mark
C1 for she has used the gradient of the radius (from O to (3, 5)); the tangent is perpendicular to the radius
oe, e.g. 'she should have used the negative reciprocal'
1 mark
B1 for \(-\frac{3}{5}\)
oe, e.g. −0.6

Question 3

3 marks

The point A has coordinates (8.4, 5.2).
The circle C has equation \(x^2 + y^2 = 36\)

Work out the shortest distance from A to the circle C.
Give your answer correct to 3 significant figures.

Model answer

The centre of C is O(0, 0) and the radius is \(\sqrt{36} = 6\)
\(OA = \sqrt{8.4^2 + 5.2^2} = \sqrt{97.6} = 9.8792...\)
Shortest distance \(= 9.8792... - 6 = 3.8792...\)
Shortest distance = 3.88 (3 s.f.)

Explanation

The shortest route from a point outside a circle to the circle is along the line towards the centre. So find the distance from A to the centre using Pythagoras, then subtract the radius. Remember that the radius is \(\sqrt{36} = 6\), not 36.

Mark scheme

1 mark
M1 for distance from O to A \(= \sqrt{8.4^2 + 5.2^2}\) (= 9.879...)
oe, e.g. \(\sqrt{97.6}\)
1 mark
M1 for radius = 6 and subtracting it from their OA
1 mark
A1 for 3.88
awrt 3.88 (3.8792...); correct answer with no working scores full marks

Question 4

5 marks

The point P(3, 6) lies on the circle \(x^2 + y^2 = 45\)
The tangent to the circle at P crosses the x-axis at A and the y-axis at B.
O is the origin.

Work out the area of triangle OAB.

Model answer

Gradient of OP \(= \dfrac{6}{3} = 2\)
Gradient of tangent \(= -\dfrac{1}{2}\)
\(y - 6 = -\frac{1}{2}(x - 3)\)
\(y = -\frac{1}{2}x + 7.5\)
B: \(x = 0\) gives \(y = 7.5\), so B = (0, 7.5)
A: \(y = 0\) gives \(\frac{1}{2}x = 7.5\), so \(x = 15\) and A = (15, 0)
Area of triangle OAB \(= \frac{1}{2} \times 15 \times 7.5 = 56.25\)

Explanation

Find the tangent's equation first: it is perpendicular to the radius OP. The triangle OAB has a right angle at O, so its base and height are the distances where the tangent crosses the axes. A common error is to use the gradient of OP (2) for the tangent.

Mark scheme

1 mark
P1 for gradient of OP \(= \frac{6}{3} = 2\)
oe; correct answer with no working scores full marks
1 mark
P1 for gradient of tangent \(= -\frac{1}{2}\)
ft their gradient of OP
1 mark
P1 for a correct equation of the tangent, e.g. \(y - 6 = -\frac{1}{2}(x - 3)\) or \(y = -\frac{1}{2}x + 7.5\)
oe, e.g. \(x + 2y = 15\)
1 mark
P1 for A = (15, 0) and B = (0, 7.5)
or OA = 15 and OB = 7.5; ft their equation
1 mark
A1 for 56.25
oe, e.g. \(\frac{225}{4}\); accept 56.3

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