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A-Level · Pure Mathematics

P3 Pure: Coordinate Geometry (Lines and Circles)

EDEXCEL 9MA0 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The points A(1, 2) and B(5, 10) are shown on a coordinate grid.

(a)Calculate the gradient of the line AB.(2)

(b)Find an equation of the line AB, giving your answer in the form y = mx + c.(2)

(c)The line l is parallel to AB and passes through the point (0, -3). Find an equation of l.(2)

(Total for Question 1 is 6 marks)

The points A(-3, -1) and B(9, 7) are joined by a straight line. The point P lies on AB such that AP : PB = 1 : 3.

(a)Find the coordinates of P.(3)

(b)Find the exact distance AP.(2)

(Total for Question 2 is 5 marks)

The points C(-2, 3) and D(4, -5) are shown on a coordinate grid.

(a)Find the midpoint M of CD.(2)

(b)Find the gradient of CD.(2)

(c)Find an equation of the perpendicular bisector of CD, giving your answer in the form ax + by + c = 0 where a, b and c are integers.(3)

(Total for Question 3 is 7 marks)

A circle has centre (2, -1) and passes through the point (5, 3).

(a)Calculate the radius of the circle.(2)

(b)Write down an equation of the circle.(2)

(c)Determine, showing your working, whether the point (7, -4) lies inside, on, or outside the circle.(3)

(Total for Question 4 is 7 marks)

A(-1, 6) and B(7, -2) are the endpoints of a diameter of circle C.

(a)Find the centre of C.(2)

(b)Find the exact radius of C.(3)

(c)Write down an equation of circle C.(2)

(Total for Question 5 is 7 marks)

The points A(1, 1), B(5, 3) and C(7, -1) form a triangle.

(a)Show that angle ABC = 90 degrees.(3)

(b)Find the exact lengths of AB and BC.(3)

(c)Hence find the area of triangle ABC.(2)

(Total for Question 6 is 8 marks)

A circle has equation (x-4)² + (y-1)² = 20. The point P(2, 5) lies on the circle.

(a)Verify that P lies on the circle.(2)

(b)Find an equation of the tangent to the circle at P, giving your answer in the form ax + by + c = 0.(4)

(c)The tangent found in part (b) meets the x-axis at A and the y-axis at B. Hence find the area of triangle OAB, where O is the origin.(3)

(Total for Question 7 is 9 marks)

The points A(1, 7) and B(9, 3) lie on a circle C. The centre of C also lies on the line with equation y = x - 1.

(a)Find an equation of the perpendicular bisector of AB.(4)

(b)Hence find an equation of circle C.(4)

(Total for Question 8 is 8 marks)

A circle C has equation (x-1)² + (y-1)² = 50. The line l has equation y = x + 6.

(a)State the coordinates of the centre and the exact radius of C.(2)

(b)Find the perpendicular distance from the centre of C to the line l.(3)

(c)Hence find the length of the chord cut from C by the line l, giving your answer in the form ksqrt(2) where k is an integer.(3)

(Total for Question 9 is 8 marks)

A circle has equation x² + y² - 6x + 4y - 37 = 0.

(a)Find the centre and the exact radius of the circle.(4)

(b)The line l has equation y = x + k, where k is a constant. Given that l is a tangent to the circle, find the two possible values of k.(6)

(Total for Question 10 is 10 marks)

A circle C has equation x² + y² = 20. The point A has coordinates (6, 2), which lies outside C.

(a)Find the exact length of the tangent from A to the circle.(3)

(b)Find the equations of the two tangents from A to the circle, giving your answers in the form y = mx + c.(7)

(Total for Question 11 is 10 marks)

A circle has centre O(0, 0) and radius 4, so that the points A(-4, 0) and B(4, 0) lie on the circle and AB is a diameter. The point P(a, b), where b > 0, also lies on the circle.

(a)Show that angle APB = 90 degrees for any point P on the circle other than A or B.(5)

(b)Given further that the tangent to the circle at P has gradient -3/4, and that a > 0, find the coordinates of P.(5)

(Total for Question 12 is 10 marks)

Mark scheme · P3 Pure: Coordinate Geometry (Lines and Circles)

Question 1

(a) M1 uses (y2 - y1)/(x2 - x1) with correct values, e.g. (10 - 2)/(5 - 1)
(a) A1 gradient = 2, cao
(a) Answer: Gradient = 2
(b) M1 uses y - 2 = 2(x - 1) or substitutes a point into y = 2x + c
(b) A1 y = 2x oe, cao
(b) Answer: y = 2x
(c) M1 identifies gradient of l is 2 (parallel lines have equal gradient)
(c) A1 y = 2x - 3 cao
(c) Answer: y = 2x - 3

Question 2

(a) M1 finds vector AB = (9 - (-3), 7 - (-1)) = (12, 8)
(a) M1 uses P = A + (1/4)AB (since AP : PB = 1 : 3 means AP is 1/4 of AB)
(a) A1 P = (0, 1) cao
(a) Answer: P = (0, 1)
(b) M1 uses AP = √(0 - (-3))² + (1 - (-1))²
(b) A1 AP = √13 exact, oe awrt 3.61
(b) Answer: AP = √13 (awrt 3.61)

Question 3

(a) M1 uses ((x1+x2)/2, (y1+y2)/2)
(a) A1 M = (1, -1) cao
(a) Answer: M = (1, -1)
(b) M1 uses (-5 - 3)/(4 - (-2))
(b) A1 gradient = -4/3 cao
(b) Answer: Gradient = -4/3
(c) M1 uses perpendicular gradient = 3/4 (negative reciprocal of -4/3)
(c) M1 forms y - (-1) = (3/4)(x - 1) using M = (1, -1)
(c) A1 3x - 4y - 7 = 0 oe with integer coefficients, cao
(c) Answer: 3x - 4y - 7 = 0

Question 4

(a) M1 uses r = √(5-2)² + (3-(-1))²
(a) A1 r = 5 cao
(a) Answer: r = 5
(b) M1 uses (x-a)² + (y-b)² = r² with centre (2,-1)
(b) A1 (x-2)² + (y+1)² = 25 cao
(b) Answer: (x-2)² + (y+1)² = 25
(c) M1 substitutes (7,-4) into (x-2)² + (y+1)², i.e. (7-2)² + (-4+1)²
(c) M1 compares result to r² = 25
(c) A1 34 > 25 so the point lies outside the circle, cao
(c) Answer: Outside the circle (34 > 25)

Question 5

(a) M1 uses the midpoint formula on A and B
(a) A1 centre = (3, 2) cao
(a) Answer: Centre = (3, 2)
(b) M1 finds length AB = √(7-(-1))² + (-2-6)²
(b) M1 halves AB to obtain the radius
(b) A1 radius = 4sqrt(2) exact, oe √32
(b) Answer: radius = 4sqrt(2)
(c) M1 uses centre (3,2) and r² = (4sqrt(2))² = 32
(c) A1 (x-3)² + (y-2)² = 32 cao
(c) Answer: (x-3)² + (y-2)² = 32

Question 6

(a) M1 finds gradient AB = (3-1)/(5-1) = 1/2
(a) M1 finds gradient BC = (-1-3)/(7-5) = -2
(a) A1 product of gradients = (1/2)(-2) = -1, so AB is perpendicular to BC, hence angle ABC = 90 degrees, cso
(a) Answer: Gradients multiply to -1, so angle ABC = 90 degrees
(b) M1 AB = √(5-1)² + (3-1)²
(b) M1 BC = √(7-5)² + (-1-3)²
(b) A1 AB = BC = 2sqrt(5) exact, both correct
(b) Answer: AB = 2sqrt(5), BC = 2sqrt(5)
(c) M1 uses area = (1/2) x AB x BC since angle ABC = 90 degrees
(c) A1 area = 10 cao
(c) Answer: Area = 10

Question 7

(a) M1 substitutes (2,5) into (x-4)² + (y-1)²
(a) A1 (2-4)² + (5-1)² = 4 + 16 = 20, which equals the right-hand side, so P lies on the circle, cso
(a) Answer: 4 + 16 = 20, so P lies on the circle
(b) M1 finds gradient of radius from centre (4,1) to P(2,5): (5-1)/(2-4) = -2
(b) M1 uses perpendicular gradient of tangent = 1/2
(b) M1 forms y - 5 = (1/2)(x - 2)
(b) A1 x - 2y + 8 = 0 oe, cao
(b) Answer: x - 2y + 8 = 0
(c) M1 sets y = 0 to find A = (-8, 0)
(c) M1 sets x = 0 to find B = (0, 4)
(c) A1 area = (1/2)(8)(4) = 16 cao
(c) Answer: Area = 16

Question 8

(a) M1 finds midpoint of AB = (5, 5)
(a) M1 finds gradient of AB = (3-7)/(9-1) = -1/2
(a) M1 uses perpendicular gradient 2 with y - 5 = 2(x - 5)
(a) A1 y = 2x - 5 oe, cao
(a) Answer: y = 2x - 5
(b) M1 solves y = 2x - 5 and y = x - 1 simultaneously
(b) A1 centre = (4, 3)
(b) M1 finds radius using distance from centre to A or B, e.g. √(4-1)²+(3-7)²
(b) A1 (x-4)² + (y-3)² = 25 cao
(b) Answer: (x-4)² + (y-3)² = 25

Question 9

(a) B1 centre = (1, 1)
(a) B1 radius = √50 = 5sqrt(2), exact
(a) Answer: Centre (1, 1), radius 5sqrt(2)
(b) M1 rearranges l into the form x - y + 6 = 0
(b) M1 applies distance = |1 - 1 + 6| / √1² + (-1)²
(b) A1 distance = 3sqrt(2) exact, oe awrt 4.24
(b) Answer: Distance = 3sqrt(2) (awrt 4.24)
(c) M1 uses r² - d² = 50 - 18 = 32
(c) M1 takes the square root and doubles: 2sqrt(32)
(c) A1 chord length = 8sqrt(2) cao
(c) Answer: Chord length = 8sqrt(2)

Question 10

(a) M1 completes the square on the x terms: x² - 6x = (x-3)² - 9
(a) M1 completes the square on the y terms: y² + 4y = (y+2)² - 4
(a) A1 centre = (3, -2)
(a) A1 radius = √50 = 5sqrt(2), exact
(a) Answer: Centre (3, -2), radius 5sqrt(2)
(b) M1 substitutes y = x + k into (x-3)² + (y+2)² = 50
(b) A1 forms the correct quadratic in x: 2x² + (2k-2)x + (k²+4k-37) = 0
(b) M1 uses the tangency condition b² - 4ac = 0
(b) A1 simplifies to k² + 10k - 75 = 0 oe
(b) A1 k = 5
(b) A1 k = -15
(b) Answer: k = 5 or k = -15

Question 11

(a) M1 finds OA² = 6² + 2² = 40, where O is the centre of C
(a) M1 uses tangent length = √OA² - r²
(a) A1 tangent length = 2sqrt(5) exact, oe √20
(a) Answer: Tangent length = 2sqrt(5)
(b) M1 sets up a line through A with gradient m: y - 2 = m(x - 6), i.e. c = 2 - 6m
(b) M1 substitutes y = mx + c into x² + y² = 20 to form (1+m²)x² + 2mcx + (c²-20) = 0
(b) M1 applies the tangency condition b² - 4ac = 0 to this quadratic
(b) A1 obtains the correct equation in m: 2m² - 3m - 2 = 0 oe
(b) A1 m = 2
(b) A1 m = -1/2
(b) A1 tangents are y = 2x - 10 and y = -(1/2)x + 5, both correct, ft from their m values
(b) Answer: y = 2x - 10 and y = -(1/2)x + 5

Question 12

(a) M1 finds gradient of AP = b/(a+4)
(a) M1 finds gradient of BP = b/(a-4)
(a) M1 forms the product of gradients = b²/(a²-16)
(a) A1 uses a² + b² = 16 (P lies on the circle) to write a² - 16 = -b²
(a) A1 product of gradients = b²/(-b²) = -1, so AP is perpendicular to BP, hence angle APB = 90 degrees, cso
(a) Answer: Gradients of AP and BP multiply to -1, so angle APB = 90 degrees
(b) M1 states gradient of radius OP = b/a
(b) M1 uses the tangent-radius perpendicularity: (b/a) x (-3/4) = -1, giving b/a = 4/3
(b) M1 substitutes b = (4/3)a into a² + b² = 16
(b) A1 a = 12/5, oe 2.4
(b) A1 b = 16/5, oe 3.2
(b) Answer: P = (12/5, 16/5)

Pure: Coordinate Geometry (Lines and Circles) - Worksheets, Questions and Revision