Circles - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Circles for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
2–8 marks each
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18 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

3 marks

The circle \(C\) has centre \((-2, 3)\) and passes through the point \(P(3, 7)\).

Find the exact \(x\)-coordinates of the points where \(C\) meets the \(x\)-axis. Give your answers in the form \(a \pm b\sqrt{2}\), where \(a\) and \(b\) are integers.

Model answer

\(r^2 = (3 - (-2))^2 + (7 - 3)^2 = 25 + 16 = 41\)
\(C: (x + 2)^2 + (y - 3)^2 = 41\)
On the \(x\)-axis, \(y = 0\):
\((x + 2)^2 + 9 = 41\)
\((x + 2)^2 = 32\)
\(x = -2 \pm \sqrt{32} = -2 \pm 4\sqrt{2}\)

Explanation

The radius is the distance from the centre to \(P\), found by Pythagoras, which gives the equation \((x + 2)^2 + (y - 3)^2 = 41\). Points on the \(x\)-axis have \(y = 0\), so \((x + 2)^2 = 32\) and \(x = -2 \pm 4\sqrt{2}\), with the surd simplified to give the form asked for. Common errors are substituting \(x = 0\) instead of \(y = 0\), sign slips when finding \(r^2\), and giving only decimal answers.

Mark scheme

1 mark
M1 Finds \(r^2\) (or \(r\)) using \(P\): \(r^2 = (3 - (-2))^2 + (7 - 3)^2 = 41\), or writes \((x + 2)^2 + (y - 3)^2 = 41\); or a correct final answer of \(x = -2 \pm 4\sqrt{2}\) with no working
Correct answers of \(x = -2 + 4\sqrt{2}\) and \(x = -2 - 4\sqrt{2}\) with no working score 3 marks. Correct decimals only (\(x = 3.66\) and \(x = -7.66\)) with no exact values score M1M1A0. Accept the expanded form \(x^2 + y^2 + 4x - 6y - 28 = 0\).
1 mark
M1 Substitutes \(y = 0\) into their equation \((x + 2)^2 + (y - 3)^2 = r^2\) (or its expanded form) and solves for \(x\), e.g. \((x + 2)^2 = 41 - 9\), or by the formula for \(x^2 + 4x - 28 = 0\)
Independent of the first M mark, but the equation must be of the form \((x + 2)^2 + (y - 3)^2 = k\) with \(k > 9\). Substituting \(x = 0\) is M0. Allow one slip.
1 mark
A1 \(x = -2 + 4\sqrt{2}\) and \(x = -2 - 4\sqrt{2}\)
Needs both M marks. Both values needed; accept \(-2 \pm 4\sqrt{2}\). \(-2 \pm \sqrt{32}\) as the final answer is A0 (not in the form asked). Isw after the correct exact answers.

Question 2

5 marks

The equation
\[x^2 + y^2 - 4kx + 2y + 5k = 0\]
where \(k\) is a constant, represents a circle.

Find the set of possible values of \(k\), giving your answer in set notation.

Model answer

\((x - 2k)^2 - 4k^2 + (y + 1)^2 - 1 + 5k = 0\)
\((x - 2k)^2 + (y + 1)^2 = 4k^2 - 5k + 1\)
For a circle, \(r^2 > 0\): \(4k^2 - 5k + 1 > 0\)
\((4k - 1)(k - 1) > 0\)
Critical values \(k = \dfrac{1}{4}\), \(k = 1\)
\(\{k : k < \tfrac{1}{4}\} \cup \{k : k > 1\}\)

Explanation

Completing the square gives \((x - 2k)^2 + (y + 1)^2 = 4k^2 - 5k + 1\), and the right-hand side is \(r^2\), which must be positive for the equation to be a circle. Solving \(4k^2 - 5k + 1 > 0\) gives the outside region of the critical values \(\tfrac{1}{4}\) and 1. Common errors are subtracting \(2k^2\) instead of \(4k^2\), and giving the answer without set notation or with \(\cap\).

Mark scheme

1 mark
M1 Attempts to complete the square for \(x\) and \(y\): \((x - 2k)^2 + (y + 1)^2 = \ldots\), or uses \(r^2 = f^2 + g^2 - c\) with \(f = -2k\), \(g = 1\), \(c = 5k\); or a correct final answer of \(\{k : k < \tfrac{1}{4}\} \cup \{k : k > 1\}\) with no working
An answer of \(\{k : k < \tfrac{1}{4}\} \cup \{k : k > 1\}\) with no working scores 5 marks. Allow \((x \pm 2k)^2 + (y \pm 1)^2\) for this mark.
1 mark
A1 \(r^2 = 4k^2 - 5k + 1\) (oe, e.g. \((x - 2k)^2 + (y + 1)^2 = 4k^2 + 1 - 5k\))
A common error is subtracting \(2k^2\) instead of \(4k^2\) when completing \(x^2 - 4kx\): A0.
1 mark
M1 (dM1) Sets their \(r^2 > 0\) (or \(= 0\)) and solves their three-term quadratic in \(k\) to find two critical values
Depends on the first M mark (not on the A mark). Usual rules for solving a quadratic; critical values from a calculator are acceptable.
1 mark
A1 Critical values \(k = \dfrac{1}{4}\) and \(k = 1\)
Accept 0.25.
1 mark
A1 \(\{k : k < \tfrac{1}{4}\} \cup \{k : k > 1\}\)
Needs the outside region in set notation joined by \(\cup\). Accept \(\{k : k < 0.25\} \cup \{k : k > 1\}\) or \(\{k \in \mathbb{R} : k < \tfrac{1}{4} \text{ or } k > 1\}\). Interval notation, e.g. \((-\infty, \tfrac{1}{4}) \cup (1, \infty)\), is not set notation here: A0. \(k < \tfrac{1}{4}\) or \(k > 1\) without set notation is A0; use of \(\cap\) or 'and' is A0; \(\leqslant\) or \(\geqslant\) (a circle of radius 0 is not a circle) is A0.

Question 3

6 marks

The circle \(C\) has equation
\[x^2 + y^2 + 4x - 6y + k = 0\]
where \(k\) is a constant. The line \(l\) with equation \(2x + y + 11 = 0\) is a tangent to \(C\) at the point \(T\).

Find the value of \(k\) and the coordinates of \(T\).

Model answer

\(l\): \(y = -2x - 11\)
\(x^2 + (-2x - 11)^2 + 4x - 6(-2x - 11) + k = 0\)
\(x^2 + 4x^2 + 44x + 121 + 4x + 12x + 66 + k = 0\)
\(5x^2 + 60x + 187 + k = 0\)
For a tangent there is a repeated root: \(b^2 - 4ac = 0\)
\(3600 - 20(187 + k) = 0\), so \(187 + k = 180\), \(k = -7\)
\(5x^2 + 60x + 180 = 0 \Rightarrow x^2 + 12x + 36 = 0 \Rightarrow (x + 6)^2 = 0\), so \(x = -6\)
\(y = -2(-6) - 11 = 1\)
\(k = -7\) and \(T(-6, 1)\)

Explanation

Substituting the line into the circle gives a quadratic, and the line is a tangent exactly when this quadratic has a repeated root, so its discriminant is zero; this gives \(k = -7\), and the repeated root gives \(T\). The geometric method is equally valid: the radius from the centre \((-2, 3)\) to \(T\) is perpendicular to \(l\), which gives \(T(-6, 1)\), then \(r^2 = 20\) and \(k\). Common errors are slips in expanding \((-2x - 11)^2\) and using \(b^2 - 4ac > 0\) instead of \(b^2 - 4ac = 0\).

Mark scheme

1 mark
M1 Substitutes \(y = -2x - 11\) (or \(x = -\dfrac{y + 11}{2}\)) into the equation of \(C\) to form a quadratic equation in one variable; correct final answers \(k = -7\) and \(T(-6, 1)\) with no working score 2 marks only (special case)
Max 2 if no working is shown. Correct answers \(k = -7\) and \(T(-6, 1)\) with no working score 2 marks, as a special case (the A1 for \(k\) and the A1 for \(T\)): the paper's rubric says that answers without working may not gain full credit. Alternative method (geometry): the centre is \((-2, 3)\), and the radius to \(T\) is perpendicular to \(l\), so it lies on \(y - 3 = \dfrac{1}{2}(x + 2)\) (M1 A1 for this line, \(y = \dfrac{1}{2}x + 4\)); solving with \(l\) gives \(T(-6, 1)\) (M1 A1); then \(r^2 = (-2 + 6)^2 + (3 - 1)^2 = 20\) and \(k = (-2)^2 + 3^2 - 20\) (M1), so \(k = -7\) (A1).
1 mark
A1 \(5x^2 + 60x + 187 + k = 0\) (oe, e.g. \(5y^2 - 10y + 33 + 4k = 0\))
1 mark
M1 Uses \(b^2 - 4ac = 0\) for a tangent (a repeated root): \(60^2 - 4 \times 5 \times (187 + k) = 0\)
Using \(b^2 - 4ac > 0\) or \(< 0\) is M0.
1 mark
A1 \(k = -7\)
1 mark
M1 (dM1) Solves their quadratic with their value of \(k\) (the repeated root \(x = -\dfrac{b}{2a}\)) and substitutes into \(l\) to find \(y\)
Depends on the M mark for the discriminant.
1 mark
A1 \(T(-6, 1)\)

Question 4

8 marks

In this question you must show all stages of your working.

The triangle \(OAB\) has vertices \(O(0, 0)\), \(A(12, 0)\) and \(B(0, 5)\). The circle \(C\) has centre \((r, r)\) and radius \(r\), where \(r\) is a positive constant, so \(C\) touches both coordinate axes.

Given that \(C\) also touches the side \(AB\) of the triangle, and that \(C\) lies inside the triangle, find the value of \(r\).

Find also the exact coordinates of the point where \(C\) touches \(AB\).

Model answer

\(AB\): gradient \(= \dfrac{5 - 0}{0 - 12} = -\dfrac{5}{12}\), so \(y = 5 - \dfrac{5}{12}x\), i.e. \(5x + 12y = 60\)
\(C\): \((x - r)^2 + (y - r)^2 = r^2\)
Substitute \(y = \dfrac{60 - 5x}{12}\) and multiply by 144:
\(144(x - r)^2 + (60 - 5x - 12r)^2 = 144r^2\)
\(169x^2 - (168r + 600)x + (60 - 12r)^2 = 0\)
\(C\) touches \(AB\), so this has a repeated root: \((168r + 600)^2 - 4 \times 169 \times (60 - 12r)^2 = 0\)
\((168r + 600)^2 = \left(26(60 - 12r)\right)^2\)
\(168r + 600 = 1560 - 312r\) or \(168r + 600 = -(1560 - 312r)\)
\(r = 2\) or \(r = 15\)
The circle with \(r = 15\) touches the axes at \((15, 0)\) and \((0, 15)\), outside the triangle, so \(r = 2\).
With \(r = 2\): \(169x^2 - 936x + 1296 = 0\), i.e. \((13x - 36)^2 = 0\), so \(x = \dfrac{36}{13}\)
\(y = \dfrac{60 - 5 \times \frac{36}{13}}{12} = \dfrac{50}{13}\)
\(r = 2\), and \(C\) touches \(AB\) at \(\left(\dfrac{36}{13}, \dfrac{50}{13}\right)\)

Explanation

The circle with centre \((r, r)\) and radius \(r\) already touches both axes, so the only condition left is that it touches \(AB\): substituting the equation of \(AB\) into the equation of the circle gives a quadratic that must have a repeated root, so its discriminant is zero. This gives \(r = 2\) or \(r = 15\), and \(r = 15\) is rejected because that circle lies outside the triangle; the repeated root then gives the point of contact. Using equal tangent lengths, \((12 - r) + (5 - r) = 13\), is a quicker valid method; common errors are slips in the expansion and not rejecting \(r = 15\).

Mark scheme

1 mark
M1 Finds an equation of the line \(AB\): gradient \(-\dfrac{5}{12}\), \(5x + 12y = 60\) (oe). An answer with no working scores 0 marks.
An answer of \(r = 2\) and \(\left(\dfrac{36}{13}, \dfrac{50}{13}\right)\) with no working scores 0 marks: the question demands all stages of working. Alternative method (equal tangents): \(AB = 13\) by Pythagoras (M1); the two tangents from \(A\) have length \(12 - r\) and the two from \(B\) have length \(5 - r\), so \((12 - r) + (5 - r) = 13\) (M1 A1); \(r = 2\) (M1 A1); the point of contact is 10 from \(A\) along \(AB\), i.e. \(A + \dfrac{10}{13}(B - A)\) (M1), giving \(\dfrac{36}{13}\) (A1) and \(\dfrac{50}{13}\) (A1). The foot of the perpendicular from \((r, r)\) to \(AB\) may also be used for the point of contact.
1 mark
M1 Substitutes the equation of \(AB\) into \((x - r)^2 + (y - r)^2 = r^2\) to form a quadratic in \(x\) (or \(y\)) with coefficients in terms of \(r\)
Independent of the first M mark. The circle equation \(x^2 + y^2 - 2rx - 2ry + r^2 = 0\) may be used.
1 mark
A1 \(169x^2 - (168r + 600)x + (60 - 12r)^2 = 0\) (oe, e.g. in \(y\): \(169y^2 - (1440 - 70r)y + (60 - 5r)^2 = 0\))
Any correct quadratic, not necessarily simplified, e.g. \(144(x - r)^2 + (60 - 5x - 12r)^2 = 144r^2\) expanded and collected.
1 mark
M1 (dM1) Uses \(b^2 - 4ac = 0\) (a repeated root, as \(C\) touches \(AB\)) to form an equation in \(r\) and solves it
Depends on the second M mark (not the point just before).
1 mark
A1 \(r = 2\), with \(r = 15\) rejected because that circle does not lie inside the triangle (it touches the axes at \((15, 0)\) and \((0, 15)\))
The rejection with a reason is needed, e.g. \(r < 5\) since \(C\) touches the side \(OB\), of length 5. Giving \(r = 15\) as well scores A0.
1 mark
M1 (dM1) Uses their \(r\) in their quadratic and finds the repeated root \(x = -\dfrac{b}{2a}\) (or uses the perpendicular from \((r, r)\) to \(AB\))
Depends on the M mark for the discriminant.
1 mark
A1 \(x = \dfrac{36}{13}\)
Exact; 2.77 scores A0.
1 mark
A1 \(y = \dfrac{50}{13}\), so the point is \(\left(\dfrac{36}{13}, \dfrac{50}{13}\right)\)
Exact; 3.85 scores A0.

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