Circles in Coordinate Geometry

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More questions on this topic: AS Pure: Co-ordinate Geometry (114) A2 Pure: Co-ordinate Geometry (77)

June 2025 Paper 1 Q22

AQACurrent spec6 marksCircles in Coordinate Geometry Simultaneous Equations

22 A circle has centre \((2, 3)\) and radius 5

A straight line passes through \((3, 7)\) and has a gradient of 2

Work out the coordinates of the two points where the circle and the straight line intersect. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(y = 2x + 1\)	B1	oe equation of straight line eg \(y - 7 = 2(x - 3)\)
\((x - 2)^2 + (y - 3)^2 = 5^2\)	B1	oe eg \(= 25\) or brackets expanded correctly
Substitutes their \((2x + 1)\) into their equation of a circle in the form \((x - a)^2 + (y - b)^2 =\) their \(5^2\) where \(a\) and \(b \ne 0\)	M1	eg \((x - 2)^2 + ((2x + 1) - 3)^2 = 5^2\) or \(x^2 - 4x + 4 + (2x + 1)^2 - 6(2x + 1) + 9 = 25\)
\(5x^2 - 12x - 17\ (= 0)\)	A1	condone \(5x^2 - 12x = 17\) if completing the square
\((5x - 17)(x + 1)\ (= 0)\) or \(\dfrac{12 \pm \sqrt{144 + 340}}{10}\) or \(5\left[\left(x - \dfrac{6}{5}\right)^2 - \dfrac{121}{25}\right]\)	M1dep	dep on B2M1A1
\((-1, -1)\) and \((3.4, 7.8)\)	A1	oe
Alternative method 2
\(x = \dfrac{y - 1}{2}\)	B1	oe equation of straight line eg \(y - 7 = 2(x - 3)\)
\((x - 2)^2 + (y - 3)^2 = 5^2\)	B1	oe eg \(= 25\) or brackets expanded correctly
Substitutes their \(\dfrac{y - 1}{2}\) into their equation of a circle in the form \((x - a)^2 + (y - b)^2 =\) their \(5^2\) where \(a\) and \(b \ne 0\)	M1	eg \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2 + (y - 3)^2 = 5^2\) or \(\left(\dfrac{y - 1}{2}\right)^2 - 4\left(\dfrac{y - 1}{2}\right) + 4 + y^2 - 6y + 9 = 25\)
\(5y^2 - 34y - 39\ (= 0)\)	A1	condone \(5y^2 - 34y = 39\) if completing the square
\((5y - 39)(y + 1)\ (= 0)\) or \(\dfrac{34 \pm \sqrt{1156 + 780}}{10}\) or \(5\left[\left(y - \dfrac{17}{5}\right)^2 - \dfrac{484}{25}\right]\)	M1dep	dep on B2M1A1 would imply B2M1A1
\((-1, -1)\) and \((3.4, 7.8)\) or \(\left(\dfrac{17}{5}, \dfrac{39}{5}\right)\)	A1

Additional guidance

First M mark can imply the B marks if correct
Expanding the circle equation and then substituting in can gain marks as an oe but won’t if the circle equation has been expanded incorrectly
Condone \((5x - 17)(5x + 5)\ (= 0)\) and \((5y - 39)(5y + 5)\ (= 0)\) for the fifth mark
Correct answer from Trial and Improvement	B2M2A2

June 2024 Paper 2 Q20

AQACurrent spec5 marksCircles in Coordinate Geometry Linear Graphs

20 \(AB\) is the tangent to the circle \(\quad (x - 3)^2 + (y - 4)^2 = 29 \quad\) at the point \((5, 9)\)

Circle with centre marked, tangent at (5, 9) meeting the y-axis at A and the x-axis at B; O is the origin

Work out the area of triangle \(AOB\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{9 - 4}{5 - 3}\) or \(\dfrac{5}{2}\)	M1	oe gradient of radius through (5, 9)
\(-1 \div\) their \(\dfrac{5}{2}\) or \(-\dfrac{2}{5}\)	M1	oe gradient of \(AB\) may be seen in equation of \(AB\) eg \(y = -\dfrac{2}{5}x + c\) their \(\dfrac{5}{2}\) cannot be \(\sqrt{29}\) or 29 gradient of \(AB = -\dfrac{2}{5}\) with no incorrect working is M1M1
\(\dfrac{y - 9}{0 - 5} =\) their \(-\dfrac{2}{5}\) or \((y =)\ 11\) or \(\dfrac{9 - 0}{5 - x} =\) their \(-\dfrac{2}{5}\) or \((x =)\ \dfrac{55}{2}\)	M1	oe eg \(y = -\dfrac{2}{5} \times 0 + 11\) or \(0 = -\dfrac{2}{5} \times x + 11\) must have scored 2nd M1 or must have scored 1st M1 and their ‘perpendicular’ gradient must be either \(1 \div\) their \(\dfrac{5}{2}\) or \(-\) their \(\dfrac{5}{2}\) \((y =)\ 11\) or \((x =)\ \dfrac{55}{2}\) may be seen on diagram
\(\dfrac{y - 9}{0 - 5} =\) their \(-\dfrac{2}{5}\) or \((y =)\ 11\) and \(\dfrac{9 - 0}{5 - x} =\) their \(-\dfrac{2}{5}\) or \((x =)\ \dfrac{55}{2}\)	M1dep	oe eg \(y = -\dfrac{2}{5} \times 0 + 11\) and \(0 = -\dfrac{2}{5} \times x + 11\) dep on 3rd M1 \((y =)\ 11\) and \((x =)\ \dfrac{55}{2}\) may be seen on diagram
\(\dfrac{605}{4}\) or \(151\dfrac{1}{4}\) or 151.25	A1	oe value

Additional guidance

Ignore conversion or simplification attempt after correct answer seen
3rd M1 and 4th M1 If using the equation of \(AB\) in the form \(y = mx + c\) they must use their gradient and have either (i) a correct value for \(c\) or (ii) a correct calculation for \(c\) eg for (ii) if \(c\) should be \(-2\) for their gradient a correct calculation for \(c\) might be \(c = 1 - 3\)
After M0M0 no other marks can be awarded
\(\dfrac{9}{5}\) (gradient of radius) \(\quad -\dfrac{5}{9}\) (gradient of \(AB\)) \(y = -\dfrac{5}{9}x + \dfrac{106}{9}\) (correct equation of \(AB\) for their gradient) \(y = \dfrac{106}{9} \quad x = \dfrac{106}{5}\) (two correct values for their line)	M0M1 M1M1A0
\(\dfrac{5}{2}\) (gradient of radius) \(\quad \dfrac{2}{5}\) (gradient of \(AB\)) \(y = \dfrac{2}{5}x + 7\) (correct equation of \(AB\) for their gradient) \(y = 7 \quad x = -\dfrac{35}{2}\) (two correct values for their line)	M1M0 M1M1A0

June 2024 Paper 2 Q12

AQACurrent spec6 marksCircles in Coordinate Geometry Solving Quadratics

12 The equation of a circle is \(\quad (x - 2)^2 + (y + 3)^2 = 16\)

The equation of a line is \(\quad y = 4 - x\)

The circle and line intersect at two points, \(A\) and \(B\).

(a) Show that the \(x\)-coordinates of \(A\) and \(B\) satisfy the equation \(\quad 2x^2 - 18x + 37 = 0\) [3 marks]

(b)

For \(A\), the \(x\)-coordinate and \(y\)-coordinate are both positive.

Work out the coordinates of \(A\).

Give each coordinate to 2 decimal places. [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(x^2 - 4x + 4\) or \(y^2 + 6y + 9\) or \(49 - 14x + x^2\)	M1	oe expansion eg \(x^2 - 2x - 2x + 4\) or \(y^2 + 3y + 3y + 9\) or \(49 - 7x - 7x + x^2\) may be embedded eg in an equation
\(x^2 - 4x + 4 + 49 - 14x + x^2 = 16\) or \(x^2 - 4x + 4 + 16 - 8x + x^2 + 24 - 6x + 9 = 16\)	M1dep	oe equation in \(x\) with all relevant expansions seen eg \(x^2 - 2x - 2x + 4 + 49 - 7x - 7x + x^2 = 16\)
M2 seen and \(2x^2 - 18x + 37 = 0\) with no rearrangement errors seen	A1

Additional guidance

After M1 an expression may be simplified when forming an equation eg \(x^2 - 4x + 4 + 49 - 14x + x^2\) \(2x^2 - 18x + 53 = 16\) (equation with all relevant expansions seen) \(2x^2 - 18x + 37 = 0\) (no rearrangement errors seen)	M1 M1 A1
For M1dep the oe equation cannot be \(2x^2 - 18x + 37 = 0\)

Mark scheme (b)

Answer	Mark	Comments
\(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 2 \times 37}}{2 \times 2}\) or \(\dfrac{18 \pm \sqrt{28}}{4}\) or [5.8, 5.823] or [3.17, 3.2]	M1	oe calculation or value for at least one \(x\)-coordinate eg \(\dfrac{9 \pm \sqrt{7}}{2}\) or \(\dfrac{9}{2} \pm \sqrt{\dfrac{7}{4}}\) condone \(+\) for \(\pm\) condone \(-\) for \(\pm\)
\(4 - \dfrac{18 + \sqrt{28}}{4}\) or \(4 -\) their [5.8, 5.823] or [−1.823, −1.8] or \(4 - \dfrac{18 - \sqrt{28}}{4}\) or \(4 -\) their [3.17, 3.2] or [0.8, 0.83]	M1dep	oe calculation or value for at least one \(y\)-coordinate
(3.18, 0.82)	A1	both coordinates must be to 2 dp

Additional guidance

Marks may be awarded from working in (a)
Using \(-18^2\) in the discriminant is M0 unless recovered
Substitution into equation of circle is possible for the 2nd M1 eg \((y + 3)^2 = 16 - (3.17 - 2)^2\) (1st M1 for 3.17 but no calculation or value of \(y\)) \((y =)\ \sqrt{16 - (3.17 - 2)^2} - 3\)	M1 M1

June 2024 Paper 1 Q6

AQACurrent spec3 marksCircles in Coordinate Geometry

6 Points \(A\ (-12, 1)\) and \(B\ (12, -1)\) lie on a circle.

\(AB\) is a diameter of the circle.

Work out the equation of the circle. [3 marks]

Mark scheme

Answer	Mark	Comments
\(x^2 + y^2 = 145\)	B3	B2 centre (0, 0) and (radius \(=\)) \(\sqrt{145}\) or centre (0, 0) and (radius\(^2 =\)) 145 or \((x - 0)^2 + (y - 0)^2 = 145\) or \(x^2 + y^2 = \sqrt{145}\) B1 \(x^2 + y^2 = k\) or \((x - 0)^2 + (y - 0)^2 = k\) or \((x + \ldots)^2 + (y + \ldots)^2 = 145\) or centre (0, 0) or (radius \(=\)) \(\sqrt{145}\) or (radius \(=\)) \(\sqrt{12^2 + 1^2}\) or (radius\(^2 =\)) 145 or \(12^2 + 1^2\) or (radius \(=\)) \(\dfrac{\sqrt{580}}{2}\) or (radius\(^2 =\)) \(\dfrac{580}{4}\)

Additional guidance

\(k\) is a positive number (not 0)
Penalise further incorrect working
(0, 0) needs to be stated as centre and not just shown on a diagram nor stated as a midpoint
No marks for finding diameter on its own

June 2023 Paper 1 Q8

AQACurrent spec6 marksCircles in Coordinate Geometry Simultaneous Equations

8 A circle has centre \((0, 0)\) and radius 5

A straight line has equation \(\quad 2y = x + 5\)

Work out the coordinates of the two points where the circle and straight line intersect.

Do not use trial and improvement.

You must show your working. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(x^2 + y^2 = 25\)	B1	oe equation of circle eg \(x^2 + y^2 = 5^2\) eg \((x - 0)^2 + (y - 0)^2 = 25\)
\((2y - 5)^2 + y^2\)	M1	must be \(2y - 5\) substituted into \(x^2 + y^2\)
\((4y^2 - 20y + 25) + y^2\)	M1	oe ft their \(2y - 5\) correctly expanded
\(5y^2 - 20y = 0\) or \(y^2 - 4y = 0\) or \(5y(y - 4) = 0\) or \(y(y - 4) = 0\)	M1dep	oe (dependent on B1 M2) must be fully correct (no ft)
\(y = 0\) and \(y = 4\) or \((-5, 0)\) or \((3, 4)\)	A1
\((-5, 0)\) and \((3, 4)\)	A1
Alternative method 2
\(x^2 + y^2 = 25\)	B1	oe equation of circle eg \(x^2 + y^2 = 5^2\) eg \((x - 0)^2 + (y - 0)^2 = 25\)
\(x^2 + \left(\dfrac{x + 5}{2}\right)^2\)	M1	must be \(\left(\dfrac{x + 5}{2}\right)\) substituted into \(x^2 + y^2\)
\(x^2 + \left(\dfrac{x^2 + 10x + 25}{4}\right)\) or \(4x^2 + x^2 + 10x + 25 \;(= 100)\)	M1	oe ft their \(\left(\dfrac{x + 5}{2}\right)\) correctly expanded
\(5x^2 + 10x - 75 = 0\) or \(x^2 + 2x - 15 = 0\) or \((x - 3)(x + 5) = 0\)	M1dep	oe (dependent on B1 M2) must be fully correct (no ft)
\(x = 3\) and \(x = -5\) or \((-5, 0)\) or \((3, 4)\)	A1
\((-5, 0)\) and \((3, 4)\)	A1

Additional guidance

Answer could be written as \(x = -5\), \(y = 0\) and \(x = 3\), \(y = 4\)	B1 M3 A2
Answer with no working (for M3) or from trying to draw an accurate graph or T&I	B0 M0 A0
Incorrect circle equation can still score 2 marks eg. \(x^2 + y^2 = 10\) \(x^2 + \left(\dfrac{x + 5}{2}\right)^2\) \(4x^2 + x^2 + 10x + 25 = 40\) \(5x^2 + 10x - 15 = 0\) \((x + 3)(x - 1) = 0\) \(x = -3\) and \(x = 1\) \(y = 1\) and \(y = 3\)	B0 M2 M0 A0 A0 B0 M1 M1 M0 A0 A0
Correct answer can be gained by using the incorrect substitution \(x = 5 - 2y\)	B1 M0 M1 M0 A0
Candidates that rearrange \(x^2 + y^2 = 25\) correctly and substitute into \(2y = x + 5\) will gain first M mark. They should then follow main scheme for other marks	B1 M1 …..

June 2023 Paper 2 Q6

AQACurrent spec2 marksCircles in Coordinate Geometry

6 The equation of a circle is \(\quad (x + 7)^2 + (y - 4)^2 = 36\)

Complete these statements. [2 marks]

The coordinates of the centre of the circle are ( ______ , ______ )

The radius of the circle is ______

Answer	Mark	Comments
\((-7, 4)\)	B1
6	B1	ignore units do not allow \(\sqrt{36}\) or \(\pm 6\)

June 2022 Paper 1 Q17

AQACurrent spec6 marksCircles in Coordinate Geometry

17 The point \(P\) lies on the circle \(\quad x^2 + y^2 = 16\)

The line \(OP\) is at an angle of \(60^\circ\) to the positive \(x\)-axis.

Circle centre O on x and y axes, with P on the circle in the first quadrant and angle 60° between OP and the positive x-axis

Not drawn accurately

(a) Show that the coordinates of point \(P\) are \((2, 2\sqrt{3})\) [2 marks]

(b) Work out the equation of the tangent to the circle at \(P\).

Write your answer in the form \(\quad x + ay = b \quad\) where \(a\) and \(b\) are constants. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
Radius of circle = 4	M1	4 could be seen in the solution or diagram without the word radius stated
Use of 4cos 60 and 4sin 60 and \(4 \times \dfrac{1}{2}\) and \(4 \times \dfrac{\sqrt{3}}{2}\)	A1	\(= (2, 2\sqrt{3})\) candidates could use the sine rule but it should look like this anyway
Alternative method 2
\(1 : \sqrt{3} : 2\) triangle seen or stated	M1	Pythagorean triple
\(2 : 2\sqrt{3} : 4\)	A1
Alternative method 3
\(\tan 60 = \dfrac{\text{opp}}{\text{adj}} = \dfrac{2\sqrt{3}}{2} = \sqrt{3}\) or \(\dfrac{\text{opp}}{\text{adj}} = \dfrac{2\sqrt{3}}{2} = \sqrt{3} = \tan 60\)	B1	shows that the point is on the line \(OP\)
\((2\sqrt{3})^2 + 2^2 = 12 + 4 = 16\)	B1	shows that the point lies on the circle

Additional guidance

Candidates could find one coordinate and then substitute into the circle equation to show the second coordinate
Candidates may try to use multiple alt methods – mark according to the method that gives them the best mark
It is possible to show that the \(x\) coordinate is 2 by connecting \(P\) and (4,0) hence creating an equilateral triangle (this would need to be stated). Then drop a perpendicular from \(P\) which bisects the base line showing that the \(x\) coordinate is 2

M1A1

Mark scheme (b)

Answer	Mark	Comments
(Gradient of \(OP\) =) \(\dfrac{2\sqrt{3}}{2}\) or \(= \sqrt{3}\)	M1	\(\sqrt{3}\) either from part (a) or knowing that an angle of \(60^\circ\) gives it
(Gradient of tangent =) \(\dfrac{-1}{\text{their}\sqrt{3}}\)	M1	oe \(\dfrac{-1}{\sqrt{3}}\) would imply the first M mark
\(y - 2\sqrt{3} = \dfrac{-1}{\sqrt{3}}(x - 2)\) or \(2\sqrt{3} = \dfrac{-1}{\sqrt{3}}(2) + c\)	M1dep	oe dependent on M2 already being awarded \(c = \dfrac{8\sqrt{3}}{3}\)
\(x + \sqrt{3}\,y = 8\)	A1

June 2022 Paper 2 Q13

AQACurrent spec3 marksCircles in Coordinate Geometry

13 \(A\,(-2, 5)\) and \(B\,(4, 13)\) are points on a circle.

\(AB\) is a diameter.

Work out the equation of the circle.

Give your answer in the form \(\quad (x - a)^2 + (y - b)^2 = c \quad\) where \(a\), \(b\) and \(c\) are integers. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\((x - 1)^2 + (y - 9)^2 = 25\)	B3	B2 \((x - 1)^2 + (y - 9)^2 = 5^2\) or (1, 9) and radius = 5 or (1, 9) and radius\(^2 = 5^2\) or (1, 9) and radius\(^2\) = 25 B1 \((x - 1)^2 + (y - 9)^2 = k\) or \((x - \ldots)^2 + (y - \ldots)^2 = 25\) or \((x - \ldots)^2 + (y - \ldots)^2 = 5^2\) or (1, 9) or \(\dfrac{-2 + 4}{2}\) oe and \(\dfrac{5 + 13}{2}\) oe or radius = 5 or radius\(^2 = 5^2\) or radius\(^2\) = 25
Alternative method 2 Uses perpendicular lines where \((x, y)\) is a point on the circle
\(\dfrac{y - 5}{x - -2} \times \dfrac{y - 13}{x - 4} = -1\)	M1	oe eg \((y - 5)(y - 13) = -1(x + 2)(x - 4)\)
\(y^2 - 18y + 65 + x^2 - 2x - 8 = 0\)	M1dep	oe equation of circle with brackets expanded and fractions eliminated eg \(y^2 - 18y + 65 = -x^2 + 2x + 8\)
\((x - 1)^2 + (y - 9)^2 = 25\)	A1

Additional guidance

\(a = 1 \quad b = 9 \quad c = 25\) implies \((x - 1)^2 + (y - 9)^2 = 25\)
Alt 1 (1, 9) may be implied eg \(x = 1 \quad y = 9\) or 1, 9	B1
Alt 1 \((x + 3)^2 + (y + 4)^2 = 5^2\)	B1
Alt 1 \((x - 1)^2 + (y - 9)^2 = 5\) (with no indication that radius = 5)	B1
Alt 1 \(r = 5\)	B1
Alt 1 diameter = 10	B0
\((x - 1)^2 + (y - 9)^2 = 25\) in working lines with brackets expanded on answer line	B2

November 2021 Paper 1 Q12

AQACurrent spec5 marksCircles in Coordinate Geometry Linear Graphs

12 A circle, centre \(C\ (4, -2)\), passes through the origin and point \(A\ (8, 0)\) on the \(x\)-axis.

The tangent at \(A\) is shown.

Circle with centre C (4, −2) passing through the origin O and A (8, 0) on the x-axis, with the tangent drawn at A

Not drawn accurately

(a) Work out the equation of the circle. [2 marks]

(b) Work out the equation of the tangent to the circle at \(A\). [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\((x - 4)^2 + (y + 2)^2 = 20\)	B2	B1 \((x - 4)^2 + (y + 2)^2\) or 20

Additional guidance

\((x + 4)^2 + (y - 2)^2 = 20\)
\((x - 4)^2 + (y + 2)^2 = 4^2 + (-2)^2\)
\((x - 4)^2 + (y + 2)^2 = \sqrt{20}\)
\((x - 4)^2 - (y + 2)^2 = 20\)
\((x - 4)^2 + (y - -2)^2 = 20\)
\((x - 4)^2 + (y - -2)^2 = (\sqrt{20})^2\)
ignore further working

B1
B1
B1
B1
B2
B2

Mark scheme (b)

Answer	Mark	Comments
(Gradient \(AC\) =) \(\dfrac{0 - -2}{8 - 4}\) or \(\dfrac{2}{4}\)	M1	oe
(Gradient of tangent =) negative reciprocal of their \(\dfrac{2}{4}\) or \(-2\)	M1	oe ft their gradient \(AC\) only gradient \(-2\) seen is M2
\(y = -2x + 16\)	A1	oe

Additional guidance

It is possible to find an incorrect gradient of \(AC\) and then get the second M mark for finding the negative reciprocal of this

M0M1A0

Specimen 2020 Paper 1 Q11

AQACurrent spec3 marksCircles in Coordinate Geometry

11 A circle, centre \(C\), touches the \(y\)-axis at the point (0, 2)

The line \(\quad y = k \quad\) intersects the circle at the points \((1, k)\) and \((5, k)\)

Circle centre C touching the y-axis at (0, 2); the horizontal line y = k cuts the circle at (1, k) and (5, k)

Not drawn accurately

Work out the equation of the circle. [3 marks]

Mark scheme

Answer	Mark	Comments
(\(x\)-coordinate of \(C\) =) \(\dfrac{5 + 1}{2}\) or 3 or (radius =) \(\dfrac{5 + 1}{2}\) or 3	M1	may be implied
(\(y\)-coordinate of \(C\) =) 2	M1	may be implied
\((x - 3)^2 + (y - 2)^2 = 9\)	A1	allow \((x - 3)^2 + (y - 2)^2 = 3^2\)

Specimen 2020 Paper 2 Q2

AQACurrent spec2 marksCircles in Coordinate Geometry

2 A circle, centre (0, 0) has circumference \(20\pi\)

Work out the equation of the circle. [2 marks]

Answer	Mark	Comments
\(x^2 + y^2 = 100\) or \(x^2 + y^2 = 10^2\)	B2	B1 radius = 10

June 2019 Paper 2 Q23

AQAOld spec6 marksCircles in Coordinate Geometry

23 A circle has centre \(C\) and equation \(\quad (x - 1)^2 + (y + 3)^2 = 25\)

\(P\,(4,\ -7)\) and \(Q\) are points on the circle.

The tangent at \(Q\) is parallel to the \(x\)-axis.
The tangents at \(P\) and \(Q\) intersect at point \(R\).

Circle with centre C below the x-axis; Q at the top of the circle with a horizontal tangent through Q and R; tangent at P (4, −7) meets it at R. Not drawn accurately

(a) Write down the coordinates of \(C\). [1 mark]

(b) Show that the equation of the tangent at \(Q\) is \(\quad y = 2\) [1 mark]

Answer	Mark	Comments
\((1,\ -3)\)	B1

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1
\(-3 + \sqrt{25}\ (= 2)\) or \(-3 + 5\ (= 2)\)	B1	oe eg \(5 - 3\ (= 2)\) or \(2 + 3 = 5\)
Alternative method 2
\((y + 3)^2 = 25\) and \(y = 2\) or \(y + 3 = 5\) and \(y = 2\) or \((2 + 3)^2 = 25\)	B1	oe eg \((1 - 1)^2 + (y + 3)^2 = 25\) and \(y = 2\)

Additional guidance

\((1,\ -3) + (0,\ 5) = (1,\ 2)\) so \(y = 2\)	B0
Allow \(-3\) + radius of 5	B1
\(2 = 0x + \mathrm{c}\) \(c = 2\) so \(y = 2\)	B0

Mark scheme (c)

Answer	Mark	Comments
Alternative method 1 Using equation \(PR\)
\(\dfrac{-7 - \text{their } -3}{4 - \text{their } 1}\) or \(-\dfrac{4}{3}\)	M1	oe grad \(PC\) their \(-3\) and their 1 from (a)
\(-1 \div \text{their } -\dfrac{4}{3}\) or \(\dfrac{3}{4}\)	M1	oe grad \(PR\) their \(-\dfrac{4}{3}\) must be a value (gradient \(PR =\)) \(\dfrac{3}{4}\) is M2
\(2 - -7 = \text{their } \dfrac{3}{4}(x - 4)\)	M1dep	oe equation \(PR\) with \(y = 2\) substituted eg \(2 = \dfrac{3}{4}x - 10\) dep on 2nd M1
16	A1ft	only ft their \(-3\) and their 1 from (a)
Alternative method 2 Using \(RC^2 = CP^2 + PR^2\) or \(PR^2 = QR^2\) with \(R\,(x,\ 2)\)
\((x - \text{their } 1)^2 + (2 - \text{their } -3)^2\) \(= (2 - \text{their } -3)^2 + (x - 4)^2 + (2 - -7)^2\)	M1	oe eg \((x - 1)^2 = (x - 4)^2 + (2 - -7)^2\) their \(-3\) and their 1 from (a)
\(x^2 - 2x + 1 + 25\) \(= 25 + x^2 - 8x + 16 + 81\)	M1dep	oe brackets expanded
\(96 = 6x\) or \(96 \div 6\)	M1dep	oe linear equation or calculation dep on M2
16	A1ft	only ft their \(-3\) and their 1 from (a)
Alternative method 3 Using equation \(CR\)
\(\dfrac{-7 - 2}{4 - \text{their } 1}\) or \(-3\)	M1	oe grad \(PQ\) their 1 from (a)
\(-1 \div \text{their } -3\) or \(\dfrac{1}{3}\)	M1	oe grad \(CR\) their \(-3\) must be a value (gradient \(CR =\)) \(\dfrac{1}{3}\) is M2
\(2 - \text{their } -3 = \text{their } \dfrac{1}{3}(x - \text{their } 1)\)	M1dep	oe equation \(CR\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on 2nd M1
16	A1ft	only ft their \(-3\) and their 1 from (a)
Alternative method 4 Using equation \(MR\) where \(M\) is the midpoint of \(PQ\)
\(\dfrac{-7 - 2}{4 - \text{their } 1}\) or \(-3\)	M1	oe grad \(PQ\) their 1 from (a)
\(-1 \div \text{their } -3\) or \(\dfrac{1}{3}\)	M1	oe grad \(MR\) their \(-3\) must be a value (gradient \(MR =\)) \(\dfrac{1}{3}\) is M2
\(\left(\dfrac{4 + \text{their } 1}{2},\ \dfrac{-7 + 2}{2}\right)\) or \((2.5,\ -2.5)\) and \(2 - \text{their } -2.5 = \text{their } \dfrac{1}{3}(x - \text{their } 2.5)\)	M1dep	oe midpoint of \(PQ\) and equation \(MR\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on 2nd M1
16	A1ft	only ft their 1 from (a)
Alternative method 5 Using equation \(MC\) where \(M\) is the midpoint of \(PQ\)
\(\left(\dfrac{4 + \text{their } 1}{2},\ \dfrac{-7 + 2}{2}\right)\) or \((2.5,\ -2.5)\)	M1	oe midpoint of \(PQ\) their 1 from (a)
\(\dfrac{\text{their } -3 - \text{their } -2.5}{\text{their } 1 - \text{their } 2.5}\) or \(\dfrac{1}{3}\)	M1dep	oe grad \(MC\)
\(2 - \text{their } -3 = \text{their } \dfrac{1}{3}(x - \text{their } 1)\) or \(2 - \text{their } -2.5 = \text{their } \dfrac{1}{3}(x - \text{their } 2.5)\)	M1dep	oe equation \(MC\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on M2
16	A1ft	only ft their \(-3\) and their 1 from (a)
Alternative method 6 Using trigonometry where \(M\) is the midpoint of \(PQ\)
\((QM =)\ \dfrac{1}{2}\sqrt{(4 - \text{their } 1)^2 + (-7 - 2)^2}\) or \(\dfrac{1}{2}\sqrt{90}\) or 4.74…	M1
\(\sin^{-1}\left(\dfrac{\text{their } 4.74\ldots}{5}\right)\) or (angle \(QCM =\)) 71.5… or 71.6	M1dep	oe angle \(QCM\)
\(\tan(\text{their } 71.5\ldots) = \dfrac{x - \text{their } 1}{5}\)	M1dep	using triangle \(QCR\)
16	A1ft	only ft their 1 from (a)

Additional guidance

Allow (16, …) to imply answer 16
Alt 1 \(-\dfrac{4}{3}x\) is M0 unless recovered
(a) \((1,\ -2)\) grad \(PC = -\dfrac{5}{3} \qquad\) grad \(PR = \dfrac{3}{5}\) Answer 19 (3rd M1 can be implied by A1ft answer)	M1M1 M1A1ft

June 2018 Paper 2 Q17

AQAOld spec5 marksCircles in Coordinate Geometry Linear Graphs

17 A circle has equation \(\qquad x^2 + y^2 = 29\)

\(P\) is the point \((-5,\ 2)\)

(a) Show that \(P\) is on the circle. [1 mark]

(b) The tangent to the circle at \(P\) intersects the \(x\)-axis at point \(Q\).

Work out the \(x\)-coordinate of \(Q\).
You must show your working. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\((-5)^2 + 2^2 = 29\)	B1	oe involving use of \(-5\) and 2 eg \((-5 - 0)^2 + (2 - 0)^2 = 29\) or \((0 - -5)^2 + (0 - 2)^2 = 29\) or \(\sqrt{(-5)^2 + 2^2} = \sqrt{29}\) or \(29 - (-5)^2 = 2^2\) or \(29 - 2^2 = (-5)^2\) or \(\sqrt{29 - (-5)^2} = 2\) or \(\sqrt{29 - 2^2} = -5\)

Additional guidance

\(25 + 4 = 29\)	B0
\(-5^2 + 2^2 = 29\)	B0
Allow 29 to be written as \(\sqrt{29}^{\,2}\)

Alt 2 diagram: P above the x-axis, with a dashed perpendicular of length 2 from P to the axis; the tangent from P meets the axis at Q, distance a from the foot of the perpendicular, and the foot is 5 from O — Mark scheme (b)

Answer	Mark	Comments
Alternative method 1 Using gradients
(gradient \(OP =\)) \(\dfrac{2 - 0}{-5 - 0}\) or \(-\dfrac{2}{5}\) or \(-0.4\)	M1	oe may be implied eg \(y = -\dfrac{2}{5}x\) or gradient of tangent \(= \dfrac{5}{2}\) (with gradient \(OP\) not seen)
(gradient tangent \(=\)) \(\dfrac{-1}{\text{their } -\dfrac{2}{5}}\) or \(\dfrac{5}{2}\) or 2.5	M1	oe correct or ft their \(-\dfrac{2}{5}\)
\(y - 2 = \text{their } \dfrac{5}{2}(x - -5)\) or \(0 - 2 = \text{their } \dfrac{5}{2}(x - -5)\) or \(2 = \text{their } \dfrac{5}{2} \times -5 + c\)	M1dep	oe dep on 2nd M1 equation of their tangent with or without substitution of \(y = 0\) implied by \(y = \dfrac{5}{2}x + \dfrac{29}{2}\) oe or \(0 = \dfrac{5}{2}x + \dfrac{29}{2}\) oe
\(-\dfrac{29}{5}\) or \(-5.8\)	A1	oe allow \(\left(-\dfrac{29}{5},\ 0\right)\) SC2 answer \(-10\) (grad tangent \(= \dfrac{2}{5}\)) SC2 answer \(-\dfrac{21}{5}\) or \(-4.2\) oe (grad tangent \(= -\dfrac{5}{2}\))
Alternative method 2 Using similar triangles (see diagram in Additional guidance)
\(\dfrac{a}{2} = \dfrac{2}{5}\)	M1	oe equation any letter
\(a = \dfrac{2}{5} \times 2\) or \(a = \dfrac{4}{5}\)	M1dep
\(-5 - \text{their } \dfrac{4}{5}\)	M1dep	dep on M2
\(-\dfrac{29}{5}\) or \(-5.8\)	A1	oe allow \(\left(-\dfrac{29}{5},\ 0\right)\) SC2 answer \(-10\) (grad tangent \(= \dfrac{2}{5}\)) SC2 answer \(-\dfrac{21}{5}\) or \(-4.2\) oe (grad tangent \(= -\dfrac{5}{2}\))

June 2018 Paper 1 Q6

AQAOld spec1 markCircles in Coordinate Geometry

6 A circle has centre \((-1,\ 2)\) and radius 5

Which of these is the equation of the circle?

Tick one box. [1 mark]

\((x + 1)^2 + (y - 2)^2 = 5\)
\((x - 1)^2 + (y + 2)^2 = 5\)
\((x + 1)^2 + (y - 2)^2 = 25\)
\((x - 1)^2 + (y + 2)^2 = 25\)

Answer	Mark	Comments
\((x + 1)^2 + (y - 2)^2 = 25\)	B1	tick in 3rd box

June 2017 Paper 2 Q22

AQAOld spec5 marksCircles in Coordinate Geometry Simultaneous Equations

22 The equation of a circle is \(\quad (x - 2)^2 + (y - 1)^2 = 16\)

The equation of a line is \(\quad y = 2x + 1\)

The circle and the line intersect at two points.

Work out the coordinates of the two points.

You must show your working.

Do not use trial and improvement. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\((x - 2)^2 + (2x + 1 - 1)^2 = 16\)	M1	oe Eliminates \(y\)
\(x^2 - 2x - 2x + 4 + 4x^2 = 16\) or \(5x^2 - 4x - 12\ (= 0)\)	M1dep	oe Expands both brackets correctly
\((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\)	M1	oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen
Alternative method 2
\(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\)	M1	oe Expands both brackets correctly
\(x^2 - 2x - 2x + 4 + (2x + 1)^2\) \(- (2x + 1) - (2x + 1) + 1 = 16\) or \(5x^2 - 4x - 12\ (= 0)\)	M1dep	oe Eliminates \(y\)
\((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\)	M1	oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen
Alternative method 3
\(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2 + (y - 1)^2 = 16\)	M1	oe Eliminates \(x\)
\(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) + 4\) \(+ y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\)	M1dep	oe Expands \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2\) and \((y - 1)^2\) correctly
\((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\)	M1	oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen
Alternative method 4
\(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\)	M1	oe Expands both brackets correctly
\(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) +\) \(4 + y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\)	M1dep	oe Eliminates \(x\)
\((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\)	M1	oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen

Additional guidance

Answers only (no valid working)	Zero
Both solutions from scale drawing	5 marks
(2, 5) is often seen without seeing any correct method	Zero
Allow one miscopy for up to M3A0A0

June 2017 Paper 1 Q10

AQAOld spec4 marksCircles in Coordinate Geometry

10 Circle C\(_1\) has centre \(L\) and equation \(\quad (x - 3)^2 + y^2 = 36\)

Circle C\(_2\) has centre \(M\) and equation \(\quad (x - h)^2 + y^2 = 64 \quad\) where \(h\) is a constant.

The circles intersect at \(N\).

\(LN\) is perpendicular to \(MN\).

Circles C1 (centre L) and C2 (centre M) on the x-axis, intersecting at N, with LN perpendicular to MN

Not drawn accurately

Work out the value of \(h\). [4 marks]

Mark scheme

Answer	Mark	Comments
(3, 0) marked or used	B1
radii 6 and 8 identified	B1	oe
\(\sqrt{(6^2 + 8^2)}\) or 6, 8, 10 triangle or 10	M1
(\(h\) =) 13 or M = (13, 0)	A1	might be seen in the working or on the diagram

Additional guidance

(3, 0) can be implied eg \(LM = h - 3\) or \(OM = 3 +\) their \(LM\)

Look on the diagram for evidence of the B marks

(\(h\) =) 13 with no working is 4 marks (\(-13\) with no working is 0 marks)

June 2016 Paper 1 Q4

AQAOld spec2 marksCircles in Coordinate Geometry

4 The equation of a circle is \(\qquad (x + 5)^2 + (y - 8)^2 = 10\)

(a) What are the coordinates of the centre of the circle?

Circle your answer. [1 mark]

\((-5, -8)\)
\((-5, 8)\)
\((5, 8)\)
\((5, -8)\)

(b) Write down the radius of the circle. [1 mark]

\((-7, +4)\)	1st B1
\(\sqrt{36}\) and 6	2nd B1

Alt 1 2nd M mark is not dependent but there must be a numerical value for grad \(OP\) to ft
grad \(OP = -0.4\) and grad tangent \(= -0.4\)	M1M0M0A0
\(\left(0,\ -\dfrac{29}{5}\right)\)	M3A0
Ignore any incorrect conversion between fraction and decimal after correct answer seen
Alt 2 diagram

Circles in Coordinate Geometry

AQA

AQA · Old spec

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