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Admissions tests / TMUA / Paper 1 / Coordinate geometry

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Coordinate geometry, set 1

Straight lines, parallel and perpendicular conditions, circles, tangents and normals, and intersections of curves with lines.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator is allowed.
  • Each question has exactly one correct answer.
  • No diagrams are provided. Where a question describes points, lines or circles, sketch the configuration yourself if it helps.
  1. 11 mark

    A straight line passes through the points (2, -1) and (6, 7). Find the equation of this line.

    1. A y = 2x - 4
    2. B y = -2x + 3
    3. C y = 2x - 5
    4. D y = (1/2)x - 2
  2. 21 mark

    Line L1 has equation 3x - y + 2 = 0. Find the equation of the line that is perpendicular to L1 and passes through the point (3, 4), giving your answer in the form ax + by + c = 0.

    1. A x + 3y - 15 = 0
    2. B 3x - y - 5 = 0
    3. C 3x + y - 13 = 0
    4. D x - 3y + 9 = 0
  3. 31 mark

    Line L1 has equation 4x - 2y + 3 = 0. Line L2 has equation kx + y - 7 = 0. Given that L1 and L2 are perpendicular, find the value of k.

    1. A k = -2
    2. B k = 1/2
    3. C k = -1/2
    4. D k = 1/4
  4. 41 mark

    A circle has equation x^2 + y^2 - 6x + 4y - 3 = 0. Find the centre and radius of this circle.

    1. A Centre (-3, 2), radius 4
    2. B Centre (3, -2), radius 16
    3. C Centre (3, -2), radius 4
    4. D Centre (3, -2), radius sqrt(10)
  5. 51 mark

    A circle has centre (1, 2) and passes through the point (4, 6). Find the equation of this circle in the form x^2 + y^2 + cx + dy + e = 0.

    1. A x^2 + y^2 + 2x + 4y - 20 = 0
    2. B x^2 + y^2 - 2x - 4y + 5 = 0
    3. C x^2 + y^2 - 2x - 4y - 20 = 0
    4. D x^2 + y^2 - 2x - 4y - 2 = 0
  6. 61 mark

    A circle has centre O and radius 13. A chord AB lies at a perpendicular distance of 5 from O. Find the length of the chord AB.

    1. A 12
    2. B 24
    3. C 16
    4. D 10
  7. 71 mark

    A circle has equation x^2 + y^2 = 20 and centre at the origin O. Find the equation of the tangent to the circle at the point P(4, 2).

    1. A y = (1/2)x
    2. B y = -(1/2)x + 4
    3. C y = -2x + 10
    4. D y = 2x - 6
  8. 81 mark

    Points A, B and C lie on a circle with centre O. The angle AOB at the centre is 136 degrees, and C lies on the major arc AB. Find the size of angle ACB.

    1. A 68 degrees
    2. B 272 degrees
    3. C 136 degrees
    4. D 112 degrees
  9. 91 mark

    AB is a diameter of a circle, and C is a point on the circle distinct from A and B. The gradient of the chord AC is 1/2. Using the fact that angle ACB is a right angle (the angle in a semicircle), find the gradient of the chord BC.

    1. A -2
    2. B 1/2
    3. C 2
    4. D -1/2
  10. 101 mark

    Points A, B, C and D lie on a circle, with C and D on the same side of the chord AB (the same segment). Angle ACB is 35 degrees. Find the size of angle ADB.

    1. A 70 degrees
    2. B 145 degrees
    3. C 35 degrees
    4. D 55 degrees
  11. 111 mark

    PQRS is a cyclic quadrilateral with vertices in that order around the circle. Angle P = (2x + 15) degrees and angle R = (x + 45) degrees. Find the value of x.

    1. A x = 100
    2. B x = 30
    3. C x = 60
    4. D x = 40
  12. 121 mark

    A tangent to a circle touches the circle at point T. A chord TS is drawn from T, and the angle between the tangent and the chord TS is 58 degrees. Point R lies on the circle in the alternate segment (the segment on the other side of TS from that angle). Find the size of angle TRS.

    1. A 32 degrees
    2. B 122 degrees
    3. C 116 degrees
    4. D 58 degrees

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. The gradient of the line through (2, -1) and (6, 7) is (7 - (-1)) / (6 - 2) = 8/4 = 2.
    2. Using the point (2, -1) in the form y - y1 = m(x - x1) gives y - (-1) = 2(x - 2), which is y + 1 = 2x - 4.
    3. Subtracting 1 from both sides gives y = 2x - 5.
    4. Therefore the equation of the line is y = 2x - 5, option C.
    • Why not A: This comes from substituting into y - y1 = m(x - x1) and then forgetting to keep the +1 on the left when rearranging, simplifying y + 1 = 2x - 4 straight to y = 2x - 4 rather than y = 2x - 5.
    • Why not B: This comes from a sign error when finding the gradient, computing the change in y as (-1) - 7 = -8 instead of 7 - (-1) = 8, giving a gradient of -8/4 = -2 instead of +2.
    • Why not D: This comes from inverting the gradient formula, computing (change in x)/(change in y) = 4/8 = 1/2 instead of (change in y)/(change in x) = 8/4 = 2.

  2. Question 2Answer: A

    1. Rearranging L1 as y = 3x + 2 shows its gradient is 3.
    2. A line perpendicular to L1 must have gradient -1/3, the negative reciprocal of 3.
    3. Using y - y1 = m(x - x1) with the point (3, 4): y - 4 = -1/3(x - 3).
    4. Multiplying through by 3 gives 3y - 12 = -(x - 3), which is 3y - 12 = -x + 3, and rearranging gives x + 3y - 15 = 0, option A.
    • Why not B: This uses the same gradient as L1 (m = 3) rather than the negative reciprocal, so it finds the line through (3, 4) that is parallel to L1, not perpendicular to it.
    • Why not C: This negates the gradient of L1 to get m = -3 but does not also take the reciprocal, so the product of the two gradients is 3 x (-3) = -9, not the -1 that perpendicularity requires.
    • Why not D: This takes the reciprocal of the gradient of L1 to get m = 1/3 but forgets to also negate it, so the product of the two gradients is 3 x (1/3) = 1, not -1.

  3. Question 3Answer: B

    1. Rearranging L1 as y = 2x + 3/2 shows its gradient is m1 = 2.
    2. Rearranging L2 as y = -kx + 7 shows its gradient is m2 = -k.
    3. Perpendicular lines satisfy m1 x m2 = -1, so 2 x (-k) = -1, which gives -2k = -1.
    4. Dividing both sides by -2 gives k = 1/2, option B.
    • Why not A: This sets the two gradients equal to each other, the condition for parallel lines (2 = -k), instead of setting their product equal to -1, the condition for perpendicular lines.
    • Why not C: This reads the coefficient of x in kx + y - 7 = 0 as the gradient of L2 directly, using m2 = k instead of correctly rearranging to y = -kx + 7 first, which gives m2 = -k.
    • Why not D: This rearranges 4x - 2y + 3 = 0 by dividing only the constant term by -2, leaving the gradient as 4 instead of correctly dividing every term to get y = 2x + 3/2, so m1 = 4 is used instead of 2.

  4. Question 4Answer: C

    1. Complete the square on the x terms: x^2 - 6x = (x - 3)^2 - 9.
    2. Complete the square on the y terms: y^2 + 4y = (y + 2)^2 - 4.
    3. Substituting back: (x-3)^2 - 9 + (y+2)^2 - 4 - 3 = 0, so (x-3)^2 + (y+2)^2 = 9 + 4 + 3 = 16.
    4. This matches the standard form (x-a)^2 + (y-b)^2 = r^2 with a = 3, b = -2 and r^2 = 16, so the centre is (3, -2) and the radius is sqrt(16) = 4, option C.
    • Why not A: This misreads the sign convention of the standard form (x-a)^2 + (y-b)^2 = r^2, taking the centre coordinates directly from the signs inside x^2 - 6x and y^2 + 4y (giving -3 and 2) instead of their negatives (giving 3 and -2).
    • Why not B: This correctly finds r^2 = 16 but forgets to take the square root, reporting the radius as 16 instead of sqrt(16) = 4.
    • Why not D: This moves the constant term -3 to the other side of the equation without flipping its sign, treating the equation as (x-3)^2 - 9 + (y+2)^2 - 4 = -3 instead of = 3, giving r^2 = 9 + 4 - 3 = 10 instead of 9 + 4 + 3 = 16.

  5. Question 5Answer: C

    1. The radius squared is the squared distance from the centre (1, 2) to the point (4, 6): r^2 = (4-1)^2 + (6-2)^2 = 9 + 16 = 25.
    2. The circle is (x-1)^2 + (y-2)^2 = 25.
    3. Expanding gives x^2 - 2x + 1 + y^2 - 4y + 4 = 25.
    4. Collecting terms and moving everything to one side: x^2 + y^2 - 2x - 4y + (1 + 4 - 25) = 0, which is x^2 + y^2 - 2x - 4y - 20 = 0, option C.
    • Why not A: This writes the circle as (x+1)^2 + (y+2)^2 = 25, using the centre coordinates with their signs unchanged inside the brackets, instead of (x-1)^2 + (y-2)^2 = 25 as the standard form (x-a)^2 + (y-b)^2 = r^2 requires for a centre of (1, 2).
    • Why not B: This correctly expands (x-1)^2 + (y-2)^2 to x^2 - 2x + 1 + y^2 - 4y + 4, but forgets to move r^2 = 25 across to combine with the +5, leaving e = 1 + 4 = 5 instead of 1 + 4 - 25 = -20.
    • Why not D: This finds r^2 by adding the coordinate differences directly, (4-1) + (6-2) = 3 + 4 = 7, instead of squaring them first, 3^2 + 4^2 = 25, so e comes out as 1 + 4 - 7 = -2 instead of 1 + 4 - 25 = -20.

  6. Question 6Answer: B

    1. The perpendicular from the centre O to the chord AB bisects the chord, so it splits AB into two equal halves and forms a right-angled triangle with the radius as the hypotenuse.
    2. The radius is 13 and the perpendicular distance is 5, so the half-chord is sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12.
    3. The full chord AB is twice this half-chord: 2 x 12 = 24, option B.
    • Why not A: This correctly finds the half-chord length using Pythagoras, sqrt(13^2 - 5^2) = sqrt(144) = 12, but reports this half-chord as the full length of AB, forgetting that the perpendicular from the centre bisects the chord into two equal halves.
    • Why not C: This treats the chord length as twice the difference between the radius and the perpendicular distance, 2 x (13 - 5) = 16, instead of applying Pythagoras' theorem to the right-angled triangle formed by the radius, the perpendicular distance and the half-chord.
    • Why not D: This takes the perpendicular distance itself, 5, to be the half-chord length and doubles it to get 10, rather than using 5 as one side of the right-angled triangle whose other side (the true half-chord) must be found via Pythagoras.

  7. Question 7Answer: C

    1. The point P(4, 2) lies on the circle since 4^2 + 2^2 = 16 + 4 = 20.
    2. The radius OP has gradient (2 - 0)/(4 - 0) = 1/2.
    3. The tangent at P is perpendicular to the radius OP, so its gradient is the negative reciprocal of 1/2, which is -2.
    4. Using y - y1 = m(x - x1) with P(4, 2): y - 2 = -2(x - 4), which gives y = -2x + 8 + 2 = -2x + 10, option C.
    • Why not A: This uses the gradient of the radius OP itself, 1/2, as the gradient of the tangent, forgetting that the tangent at a point is perpendicular to the radius at that point, not parallel to it.
    • Why not B: This negates the radius's gradient of 1/2 to get -1/2 but does not also take the reciprocal, so the product of the two gradients is 1/2 x (-1/2) = -1/4, not the -1 that perpendicularity requires.
    • Why not D: This takes the reciprocal of the radius's gradient (1/2 becomes 2) but forgets to also negate it, so the product of the two gradients is 1/2 x 2 = 1, not -1.

  8. Question 8Answer: A

    1. The angle at the centre of a circle is twice the angle at the circumference when both angles are subtended by the same arc.
    2. Because C lies on the major arc AB, angle ACB is subtended by the same (minor) arc as the given centre angle AOB.
    3. So angle ACB = 136 / 2 = 68 degrees, option A.
    • Why not B: This doubles the centre angle instead of halving it, applying the centre-circumference rule backwards: 136 x 2 = 272.
    • Why not C: This uses the angle at the centre directly as the answer, forgetting that the angle at the circumference is half the angle at the centre, not equal to it.
    • Why not D: This uses the reflex angle at the centre, 360 - 136 = 224, and halves that instead, which would only be correct if C lay on the minor arc rather than the major arc as stated.

  9. Question 9Answer: A

    1. Since AB is a diameter, the angle in a semicircle theorem gives angle ACB = 90 degrees.
    2. A right angle between two chords AC and BC means their gradients multiply to give -1.
    3. If the gradient of AC is 1/2, the gradient of BC must be the negative reciprocal of 1/2, which is -2, option A.
    • Why not B: This assumes AC and BC have the same gradient, as if they were parallel, but the right angle at C means the two chords must be perpendicular, not parallel.
    • Why not C: This takes the reciprocal of the gradient of AC (1/2 becomes 2) but forgets to also negate it, so the product of the two gradients is 1/2 x 2 = 1, not -1.
    • Why not D: This negates the gradient of AC to get -1/2 but does not also take the reciprocal, so the product of the two gradients is 1/2 x (-1/2) = -1/4, not the -1 that perpendicularity requires.

  10. Question 10Answer: C

    1. C and D lie on the same side of chord AB, so they lie in the same segment.
    2. Angles subtended at the circumference by the same arc, from points in the same segment, are equal.
    3. Angle ACB and angle ADB are both subtended by arc AB from the same segment, so angle ADB = angle ACB = 35 degrees, option C.
    • Why not A: This doubles angle ACB as though converting a circumference angle to a centre angle, but no centre is involved here: both C and D are circumference points in the same segment, subtending the same arc AB.
    • Why not B: This applies the cyclic quadrilateral rule (opposite angles sum to 180 degrees) to angle ACB and angle ADB, but these are not opposite angles of a cyclic quadrilateral; they are two circumference angles subtending the same arc from the same side, and the correct rule is that such angles are equal.
    • Why not D: This assumes AB must be a diameter and uses 90 degrees as a reference to subtract from, but the question never states AB is a diameter, and even if it were, that would only fix each of angle ACB and angle ADB individually at 90 degrees, not give a rule for combining them by subtraction.

  11. Question 11Answer: D

    1. P and R are opposite vertices of the cyclic quadrilateral PQRS, so angle P and angle R are opposite angles and must sum to 180 degrees.
    2. So (2x + 15) + (x + 45) = 180, which simplifies to 3x + 60 = 180.
    3. Subtracting 60 from both sides gives 3x = 120.
    4. Dividing by 3 gives x = 40, option D.
    • Why not A: This sets the two angles summing to 360 degrees, the angle sum of a whole quadrilateral, instead of 180 degrees, the rule specific to a pair of OPPOSITE angles in a cyclic quadrilateral.
    • Why not B: This sets angle P equal to angle R, as though opposite angles in a cyclic quadrilateral were equal like opposite sides of a parallelogram, rather than supplementary.
    • Why not C: This collects the two x-terms, 2x and x, as a single 2x rather than 3x, dropping one of the x-terms when combining like terms.

  12. Question 12Answer: D

    1. The alternate segment theorem states that the angle between a tangent and a chord at the point of contact equals the angle subtended by that chord in the alternate segment.
    2. The tangent-chord angle at T is 58 degrees, and R lies in the alternate segment relative to that angle.
    3. So angle TRS = 58 degrees, option D.
    • Why not A: This treats the tangent-chord angle as complementary to angle TRS, subtracting it from 90 degrees, as if the tangent-radius right angle applied directly here, rather than applying the alternate segment theorem.
    • Why not B: This treats angle TRS as the supplement of the tangent-chord angle, subtracting it from 180 degrees as though the two angles lay together on a straight line, rather than applying the alternate segment theorem, which makes the two angles equal.
    • Why not C: This doubles the tangent-chord angle, confusing this configuration with the rule that an angle at the centre is twice the angle at the circumference; no centre is involved in the alternate segment theorem.

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