Circles - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Circles for Edexcel A-Level Maths (Paper 2). 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.

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2–9 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 \(C\) passes through the origin \(O\), through the point \(A(10, 0)\) on the \(x\)-axis and through the point \(B(0, -4)\) on the \(y\)-axis.

Explain why \(AB\) is a diameter of \(C\), and find an equation for \(C\).

Model answer

The \(x\)-axis and \(y\)-axis are perpendicular, so angle \(AOB = 90^\circ\).
\(O\) is on \(C\), and the angle in a semicircle is a right angle, so the chord \(AB\) is a diameter.
Centre = midpoint of \(AB = \left(\dfrac{10 + 0}{2}, \dfrac{0 - 4}{2}\right) = (5, -2)\)
\(r^2 = (5 - 0)^2 + (-2 - 0)^2 = 29\)
\(C: (x - 5)^2 + (y + 2)^2 = 29\)

Explanation

Because the axes meet at right angles, the chord \(AB\) subtends \(90^\circ\) at the point \(O\) on the circle, and by the angle-in-a-semicircle property \(AB\) must be a diameter. The centre is then the midpoint \((5, -2)\) and the radius is its distance from \(O\), \(\sqrt{29}\). Common errors are an explanation that does not mention the right angle and using the length of \(AB\) as the radius.

Mark scheme

1 mark
M1 Finds the centre as the midpoint of \(AB\), \((5, -2)\), and the radius from it, e.g. \(r^2 = 5^2 + 2^2\); or substitutes \(O\), \(A\) and \(B\) into \(x^2 + y^2 + ax + by + c = 0\) to find \(c = 0\), \(a = -10\), \(b = 4\); or a correct equation \((x - 5)^2 + (y + 2)^2 = 29\) with no working
An answer of \((x - 5)^2 + (y + 2)^2 = 29\) with no working scores 2 marks; the B1 needs the explanation. Using the whole length \(AB = \sqrt{116}\) as the radius is M0.
1 mark
A1 \((x - 5)^2 + (y + 2)^2 = 29\) or \(x^2 + y^2 - 10x + 4y = 0\)
Accept any equivalent equation. Centre and radius stated without an equation is A0.
1 mark
B1 The axes are perpendicular, so angle \(AOB = 90^\circ\); \(O\) lies on \(C\), and a chord that subtends a right angle at a point on the circle is a diameter (the angle in a semicircle is a right angle), so \(AB\) is a diameter
Must refer to the right angle at \(O\) (or the perpendicular axes) and link it to the angle in a semicircle or to a diameter. 'Because O is on the circle' alone is B0. Also accept a check that the midpoint of \(AB\) is equidistant (\(\sqrt{29}\)) from \(O\), \(A\) and \(B\), so it is the centre and \(AB\) passes through it.

Question 2

5 marks

The circle \(C\) has equation
\[x^2 + y^2 + 4x - 2y - 24 = 0\]
Show that the point \(P(3, 3)\) lies on \(C\).

Find an equation of the tangent to \(C\) at \(P\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.

Model answer

\(3^2 + 3^2 + 4(3) - 2(3) - 24 = 9 + 9 + 12 - 6 - 24 = 0\), so \(P\) lies on \(C\).
\((x + 2)^2 + (y - 1)^2 = 29\), centre \((-2, 1)\)
Gradient of radius \(CP = \dfrac{3 - 1}{3 + 2} = \dfrac{2}{5}\)
Gradient of tangent \(= -\dfrac{5}{2}\)
\(y - 3 = -\dfrac{5}{2}(x - 3)\)
\(2y - 6 = -5x + 15\)
\(5x + 2y - 21 = 0\)

Explanation

The tangent at \(P\) is perpendicular to the radius \(CP\), so its gradient is the negative reciprocal of the gradient of \(CP\), \(-\dfrac{5}{2}\). Marks are for the centre and radius gradient, the perpendicular gradient, the line through \(P\) and the equation in the form asked. Common errors are using the gradient of the radius itself and leaving the answer as \(y = mx + c\).

Mark scheme

1 mark
B1 Substitutes \(x = 3\), \(y = 3\): \(9 + 9 + 12 - 6 - 24 = 0\), so \(P\) lies on \(C\); a correct tangent \(5x + 2y - 21 = 0\) with no working scores the other 4 marks
An answer of \(5x + 2y - 21 = 0\) with no working scores 4 marks; this B1 needs the substitution shown with a conclusion. Accept \((3 + 2)^2 + (3 - 1)^2 = 29\) from the completed-square form.
1 mark
M1 Finds the centre \((-2, 1)\) (by completing the square or from \((-g, -f)\)) and the gradient of the radius \(CP\): \(\dfrac{3 - 1}{3 - (-2)} = \dfrac{2}{5}\)
Allow a sign slip in the centre for this mark, e.g. \((2, -1)\) giving a gradient of 2. Alternative: implicit differentiation \(2x + 2y\dfrac{dy}{dx} + 4 - 2\dfrac{dy}{dx} = 0\) with \(x = y = 3\) scores this mark and the next for \(\dfrac{dy}{dx} = -\dfrac{5}{2}\).
1 mark
M1 (dM1) Uses the perpendicular gradient: tangent gradient \(= -\dfrac{1}{\text{their } \frac{2}{5}} = -\dfrac{5}{2}\)
Depends on the previous M mark.
1 mark
M1 Forms the equation of a line through \(P(3, 3)\) with their tangent gradient: \(y - 3 = -\dfrac{5}{2}(x - 3)\)
Their tangent gradient must be a changed gradient, not the gradient of the radius. If \(y = mx + c\) is used they must reach a value for \(c\) (\(c = \dfrac{21}{2}\)).
1 mark
A1 \(5x + 2y - 21 = 0\)
Needs all three M marks. Accept any integer multiple, e.g. \(-5x - 2y + 21 = 0\) or \(10x + 4y - 42 = 0\), and the terms in any order provided the form is \(ax + by + c = 0\). \(y = -\dfrac{5}{2}x + \dfrac{21}{2}\) is A0 (not the form asked).

Question 3

7 marks

The circle \(C\) has equation
\[x^2 + y^2 - 6x + 2ky + 25 = 0\]
where \(k\) is a constant.

Find, in terms of \(k\), the coordinates of the centre of \(C\) and the radius of \(C\).

Given that \(C\) touches the \(y\)-axis, find the possible values of \(k\).

Show that, whatever the value of \(k\), \(C\) does not meet the \(x\)-axis.

Model answer

\((x - 3)^2 - 9 + (y + k)^2 - k^2 + 25 = 0\)
\((x - 3)^2 + (y + k)^2 = k^2 - 16\)
Centre \((3, -k)\), radius \(\sqrt{k^2 - 16}\)
Touches the \(y\)-axis: the radius equals the distance 3 from the centre to the \(y\)-axis.
\(k^2 - 16 = 9 \Rightarrow k^2 = 25 \Rightarrow k = 5\) or \(k = -5\)
On the \(x\)-axis, \(y = 0\): \(x^2 - 6x + 25 = 0\)
\(b^2 - 4ac = 36 - 100 = -64 < 0\), so there are no real roots.
This does not depend on \(k\), so \(C\) never meets the \(x\)-axis.

Explanation

Completing the square gives centre \((3, -k)\) and radius \(\sqrt{k^2 - 16}\); touching the \(y\)-axis means the radius equals 3, the distance from the centre to that axis, so \(k = \pm 5\). Putting \(y = 0\) removes \(k\) entirely and leaves \(x^2 - 6x + 25 = 0\), whose discriminant is negative, so the circle never meets the \(x\)-axis. Common errors are mistakes with \(-k^2 + 25\) in the radius and checking only one value of \(k\) for the last part.

Mark scheme

1 mark
M1 Completes the square: \((x - 3)^2 + (y + k)^2 = 9 + k^2 - 25\) (allow sign slips); or correct answers of centre \((3, -k)\), radius \(\sqrt{k^2 - 16}\) and \(k = \pm 5\) with no working
An answer of centre \((3, -k)\), radius \(\sqrt{k^2 - 16}\) and \(k = 5\) or \(-5\) with no working scores 5 marks; the last two marks need the argument shown. Using \((-g, -f)\) and \(\sqrt{g^2 + f^2 - c}\) is fine.
1 mark
A1 Centre \((3, -k)\)
1 mark
A1 Radius \(\sqrt{k^2 - 16}\)
Needs the M mark. Common errors: \(\sqrt{k^2 + 16}\), \(\sqrt{k^2 - 34}\), \(k - 4\): A0.
1 mark
M1 Uses the fact that the radius equals the distance 3 from the centre to the \(y\)-axis: \(k^2 - 16 = 9\); or substitutes \(x = 0\), \(y^2 + 2ky + 25 = 0\), and sets the discriminant \(4k^2 - 100 = 0\)
Follow through their radius. Using radius \(= |k|\) here (the condition for the \(x\)-axis) is M0.
1 mark
A1 \(k = 5\) or \(k = -5\)
Both values needed.
1 mark
M1 Substitutes \(y = 0\): \(x^2 - 6x + 25 = 0\), and finds its discriminant \(36 - 100\); or compares the radius \(\sqrt{k^2 - 16}\) with the distance \(|k|\) of the centre from the \(x\)-axis
Completing the square, \((x - 3)^2 + 16 = 0\), is also fine.
1 mark
A1* \(b^2 - 4ac = -64 < 0\), so there are no real solutions and \(C\) does not meet the \(x\)-axis for any \(k\) (or: \(k^2 - 16 < k^2\), so the radius is always less than the distance of the centre from the \(x\)-axis) cso
Needs a correct reason and a conclusion that holds for every \(k\). Testing particular values of \(k\) only is A0.

Question 4

8 marks

The circle \(C\) has equation
\[x^2 + y^2 - 4x + 6y - 37 = 0\]
and centre \(K\). The line \(l\) with equation \(3x - 4y + 7 = 0\) meets \(C\) at the points \(A\) and \(B\), where the \(x\) coordinate of \(A\) is positive.

Find the coordinates of \(A\) and of \(B\).

Show that angle \(AKB\) is a right angle.

The line \(l\) divides the region inside \(C\) into two parts. Find the exact area of the smaller part, giving your answer in the form \(p\pi + q\), where \(p\) and \(q\) are rational numbers.

Model answer

\((x - 2)^2 + (y + 3)^2 = 4 + 9 + 37 = 50\): centre \(K(2, -3)\), radius \(\sqrt{50}\)
From \(l\): \(x = \dfrac{4y - 7}{3}\), so \(x - 2 = \dfrac{4y - 13}{3}\)
\(\dfrac{(4y - 13)^2}{9} + (y + 3)^2 = 50\)
\(16y^2 - 104y + 169 + 9y^2 + 54y + 81 = 450\)
\(25y^2 - 50y - 200 = 0 \Rightarrow y^2 - 2y - 8 = 0 \Rightarrow (y - 4)(y + 2) = 0\)
\(y = 4\): \(x = 3\); \(y = -2\): \(x = -5\)
\(A(3, 4)\), \(B(-5, -2)\)
Gradient of \(KA = \dfrac{4 + 3}{3 - 2} = 7\), gradient of \(KB = \dfrac{-2 + 3}{-5 - 2} = -\dfrac{1}{7}\)
\(7 \times \left(-\dfrac{1}{7}\right) = -1\), so \(KA\) is perpendicular to \(KB\): angle \(AKB = 90^\circ\).
The smaller part is the minor segment: sector \(AKB\) minus triangle \(AKB\).
Area \(= \dfrac{1}{2}(50)\left(\dfrac{\pi}{2}\right) - \dfrac{1}{2}\left(\sqrt{50}\right)^2 = \dfrac{25}{2}\pi - 25\)

Explanation

Substituting the line into the circle gives a quadratic whose roots lead to \(A(3, 4)\) and \(B(-5, -2)\), and the radii \(KA\) and \(KB\) have gradients 7 and \(-\dfrac{1}{7}\), whose product is \(-1\), so the angle at the centre is a right angle. The smaller part is the minor segment: the quarter-circle sector of area \(\dfrac{1}{2}r^2\theta = \dfrac{25}{2}\pi\) minus the right-angled triangle \(AKB\) of area 25. Common errors are sign slips in the centre, adding the triangle instead of subtracting it, and using \(\theta = 90\) in \(\dfrac{1}{2}r^2\theta\).

Mark scheme

1 mark
M1 Substitutes the equation of \(l\) into the equation of \(C\) to obtain an equation in one variable, e.g. \(x^2 + \left(\dfrac{3x + 7}{4}\right)^2 - 4x + 6\left(\dfrac{3x + 7}{4}\right) - 37 = 0\) or \(\left(\dfrac{4y - 7}{3} - 2\right)^2 + (y + 3)^2 = 50\); correct answers of \(A(3, 4)\), \(B(-5, -2)\) and \(\dfrac{25}{2}\pi - 25\) with no working score 2 marks
Correct answers of \(A(3, 4)\), \(B(-5, -2)\) and an area of \(\dfrac{25}{2}\pi - 25\) with no working score 2 marks (special case: the A mark for the two points and the final A mark), since this is a multi-step problem; the right angle must be shown with working. Allow one slip in the substitution for this mark. Using the completed-square form \((x - 2)^2 + (y + 3)^2 = 50\) is equally good.
1 mark
A1 A correct three-term quadratic: \(25x^2 + 50x - 375 = 0\), i.e. \(x^2 + 2x - 15 = 0\) (or \(25y^2 - 50y - 200 = 0\), i.e. \(y^2 - 2y - 8 = 0\))
Any correct form, simplified or not; '= 0' may be implied.
1 mark
M1 (dM1) Solves their quadratic and finds the other coordinate of each point from the equation of \(l\)
Depends on the first M mark. Roots of a correct quadratic may be written down from a calculator.
1 mark
A1 \(A(3, 4)\) and \(B(-5, -2)\)
Both points needed. Accept \(x = 3, y = 4\) and \(x = -5, y = -2\). Condone the labels \(A\) and \(B\) the wrong way round for this mark.
1 mark
M1 Finds the centre \(K(2, -3)\) and uses a valid test for a right angle at \(K\) with their \(A\) and \(B\): the gradients of \(KA\) and \(KB\) and their product; or \(KA^2 + KB^2\) compared with \(AB^2\); or the cosine rule in triangle \(AKB\)
Follow through their points. Gradients: \(KA\) has gradient \(\dfrac{4 + 3}{3 - 2} = 7\) and \(KB\) has gradient \(\dfrac{-2 + 3}{-5 - 2} = -\dfrac{1}{7}\). Pythagoras: \(KA^2 = KB^2 = 50\) (the radius squared) and \(AB^2 = 8^2 + 6^2 = 100\). A centre of \((-2, 3)\) is M0.
1 mark
A1* \(7 \times \left(-\dfrac{1}{7}\right) = -1\) (or \(KA^2 + KB^2 = 50 + 50 = 100 = AB^2\)), so \(KA\) is perpendicular to \(KB\) and angle \(AKB = 90^\circ\) cso
Needs correct points, the correct centre, the calculation and a conclusion. Stating that the angle is \(90^\circ\) with no calculation is A0.
1 mark
M1 Area of the smaller part \(=\) area of the sector \(AKB\) with angle \(\dfrac{\pi}{2}\) \(-\) area of triangle \(AKB\): \(\dfrac{1}{2}(50)\left(\dfrac{\pi}{2}\right) - \dfrac{1}{2}\left(\sqrt{50}\right)\left(\sqrt{50}\right)\)
May use the right angle even if it was not shown. Follow through their \(r^2\). The sector as a quarter of \(\pi r^2\), or \(\dfrac{90}{360} \times \pi r^2\), is fine. Sector plus triangle is M0, and \(\dfrac{1}{2}r^2\theta\) with \(\theta = 90\) is M0.
1 mark
A1 \(\dfrac{25}{2}\pi - 25\), i.e. \(p = \dfrac{25}{2}\), \(q = -25\)
Accept \(12.5\pi - 25\) or \(\dfrac{25\pi - 50}{2}\). A decimal only (14.3) is A0; isw after the exact answer. The larger part, \(\dfrac{75}{2}\pi + 25\), is A0.

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