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Admissions tests / TMUA / Paper 1 / Coordinate geometry

Demanding. 12 questions, 12 marks, about 50 minutes.

TMUA Paper 1: Coordinate geometry, set 2

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.
  • This is a demanding set: several questions need two or three linked steps, and there is less signposting than the real exam.
  1. 11 mark

    A circle has centre lying on the line x + y = 15. The circle passes through the points A(1, 6) and B(9, 2). Find the coordinates of the centre.

    1. A (5, 4)
    2. B (17, -2)
    3. C (7, 8)
    4. D (-1, 16)
  2. 21 mark

    The line y = 2x + c is tangent to the circle x^2 + y^2 = 5. Find the possible values of c.

    1. A c = 5 or c = -5
    2. B c = 5/2 or c = -5/2
    3. C c = sqrt(5) or c = -sqrt(5)
    4. D c = 5sqrt(5) or c = -5sqrt(5)
  3. 31 mark

    Find the range of values of k for which the line y = x + k intersects the circle x^2 + y^2 = 18 at two distinct points.

    1. A -6 <= k <= 6
    2. B k < -6 or k > 6
    3. C -3sqrt(2) < k < 3sqrt(2)
    4. D -6 < k < 6
  4. 41 mark

    Points A and B lie on a circle with centre O. The reflex angle AOB is 250 degrees, and C is a point on the minor arc AB. Find the size of angle ACB.

    1. A 55 degrees
    2. B 125 degrees
    3. C 250 degrees
    4. D 70 degrees
  5. 51 mark

    AB is a diameter of a circle, with A(-3, 1) and B(5, 7). Point C lies on the circle with AC = 6. Using the fact that angle ACB is a right angle (the angle in a semicircle), find the length of BC.

    1. A 4
    2. B 8
    3. C 2sqrt(34)
    4. D 2sqrt(7)
  6. 61 mark

    PQRS is a cyclic quadrilateral with vertices in that order around the circle. Angle P = (3x + 10) degrees, angle Q = (2x - 5) degrees and angle R = (x + 50) degrees. Find the size of angle S.

    1. A 55 degrees
    2. B 35 degrees
    3. C 80 degrees
    4. D 125 degrees
  7. 71 mark

    A tangent to a circle at point T makes an angle of 47 degrees with the chord TQ, measured on one side. Point R lies on the circle in the alternate segment relative to this angle, so angle TRQ = 47 degrees by the alternate segment theorem. In triangle TQR, angle RTQ = 71 degrees. Find angle TQR.

    1. A 62 degrees
    2. B 47 degrees
    3. C 118 degrees
    4. D 109 degrees
  8. 81 mark

    A circle has centre C(2, -1) and radius 5. Find the length of the tangent from the external point P(14, 4) to the circle.

    1. A 13
    2. B 8
    3. C 12
    4. D 144
  9. 91 mark

    A line passes through the point (3, 4) and is parallel to the line 4x + 3y = 5. This line crosses the x-axis at A and the y-axis at B. Find the length AB.

    1. A 5/4
    2. B 35/4
    3. C 14
    4. D 10
  10. 101 mark

    Points A, B, C and D lie on a circle, with C and D on the same side of chord AB (the same segment). Angle ACB = (4x - 15) degrees and angle ADB = (2x + 25) degrees. Given that angle AOB at the centre O, subtended by the same arc AB as angle ACB, is twice angle ACB, find the size of angle AOB.

    1. A 65 degrees
    2. B 115 degrees
    3. C 130 degrees
    4. D 50 degrees
  11. 111 mark

    A circle has equation x^2 + y^2 + 4x - 10y + k = 0 and has radius 7. Find the value of k.

    1. A 22
    2. B -20
    3. C -78
    4. D 20
  12. 121 mark

    Points A and B lie on a circle with centre O and radius r. The angle OAB between the radius OA and the chord AB is 35 degrees. Point C lies on the major arc AB. Find the size of angle ACB.

    1. A 55 degrees
    2. B 72.5 degrees
    3. C 110 degrees
    4. D 70 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 midpoint of AB is ((1+9)/2, (6+2)/2) = (5, 4).
    2. The gradient of AB is (2-6)/(9-1) = -4/8 = -1/2, so the perpendicular bisector of AB, which passes through the centre, has gradient 2.
    3. The perpendicular bisector is y - 4 = 2(x - 5), which simplifies to y = 2x - 6.
    4. The centre also lies on x + y = 15. Substituting gives x + (2x - 6) = 15, so 3x = 21 and x = 7, giving y = 8.
    5. Therefore the centre is (7, 8), option C.
    • Why not A: This assumes the centre is simply the midpoint of the chord AB, forgetting that this only holds if AB is a diameter; here the centre must instead be found from the perpendicular bisector of AB together with the given line.
    • Why not B: This uses the gradient of the chord AB itself (-1/2) for the bisecting line through its midpoint, instead of the perpendicular gradient (2), forgetting that the line from the centre to the midpoint of a chord must be perpendicular to the chord, not parallel to it.
    • Why not D: This takes the reciprocal of the chord's gradient (-1/2 becomes -2) to get the perpendicular bisector's gradient, but forgets to also change its sign, so the product of the two gradients is (-1/2) x (-2) = 1, not the -1 that perpendicularity requires.

  2. Question 2Answer: A

    1. Substituting y = 2x + c into x^2 + y^2 = 5 gives x^2 + (2x + c)^2 = 5, which expands to 5x^2 + 4cx + (c^2 - 5) = 0.
    2. A tangent line meets the circle at exactly one point, so this quadratic in x must have a repeated root, meaning its discriminant is 0.
    3. (4c)^2 - 4(5)(c^2 - 5) = 0, so 16c^2 - 20c^2 + 100 = 0, which gives -4c^2 + 100 = 0.
    4. So c^2 = 25, giving c = 5 or c = -5, option A.
    • Why not B: This squares only the c in the term (4c)^2, writing it as 4c^2 instead of 16c^2, which shrinks that term in the discriminant to a quarter of its true size and changes c^2 = 25 into c^2 = 25/4.
    • Why not C: This expands (2x + c)^2 as 4x^2 + c^2, dropping the middle term 4cx entirely, which removes the linear term from the quadratic and leads to c^2 = 5 instead of c^2 = 25.
    • Why not D: This reads the radius of x^2 + y^2 = 5 as 5 instead of sqrt(5) (forgetting to take the square root of the constant term), then uses this wrong radius in the distance-from-centre-equals-radius method.

  3. Question 3Answer: D

    1. Substituting y = x + k into x^2 + y^2 = 18 gives x^2 + (x + k)^2 = 18, which simplifies to 2x^2 + 2kx + (k^2 - 18) = 0.
    2. Two distinct intersection points require this quadratic to have two distinct real roots, so the discriminant must be strictly positive.
    3. (2k)^2 - 4(2)(k^2 - 18) > 0, so 4k^2 - 8k^2 + 144 > 0, which gives -4k^2 + 144 > 0.
    4. Dividing by -4 and reversing the inequality gives k^2 < 36, so -6 < k < 6, option D.
    • Why not A: This uses discriminant >= 0 (which includes the boundary case where the line is tangent, touching at exactly one point) instead of discriminant > 0, which is what two DISTINCT intersection points actually requires.
    • Why not B: This divides -4k^2 + 144 > 0 by -4 without reversing the direction of the inequality, landing on k^2 > 36 instead of k^2 < 36; k^2 > 36 is in fact the condition for the line to miss the circle entirely, not cut it.
    • Why not C: This expands (x + k)^2 as x^2 + k^2, dropping the middle term 2kx, which removes the linear term from the quadratic and leads to k^2 < 18 instead of k^2 < 36.

  4. Question 4Answer: B

    1. Since C lies on the minor arc AB, the arc AB not containing C is the major arc, and its central angle is the reflex angle AOB = 250 degrees.
    2. By the circle theorem, the angle at the centre is twice the angle at the circumference subtended by the same arc.
    3. So angle ACB = 250 / 2 = 125 degrees, option B.
    • Why not A: This uses the non-reflex centre angle, 360 - 250 = 110, and halves it to get 55; that would be correct if C were on the major arc, but since C is on the minor arc, angle ACB is subtended by the major arc, whose centre angle is the reflex 250 degrees.
    • Why not C: This correctly identifies that the reflex angle is the one to use, but then forgets that the angle at the centre is twice the angle at the circumference, reporting the centre angle itself instead of halving it.
    • Why not D: This misapplies the cyclic-quadrilateral rule that opposite angles sum to 180 degrees, subtracting the non-reflex centre angle from 180 (180 - 110 = 70), even though no cyclic quadrilateral is involved here.

  5. Question 5Answer: B

    1. AB = sqrt((5-(-3))^2 + (7-1)^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10.
    2. Since AB is a diameter, angle ACB = 90 degrees (the angle in a semicircle), so triangle ACB is right-angled at C with hypotenuse AB.
    3. By Pythagoras, AC^2 + BC^2 = AB^2, so 36 + BC^2 = 100, giving BC^2 = 64.
    4. So BC = 8, option B.
    • Why not A: This computes BC = AB - AC = 10 - 6 = 4, treating the three lengths as if they add linearly along a straight line instead of forming a right-angled triangle where AC^2 + BC^2 = AB^2.
    • Why not C: This computes BC = sqrt(AB^2 + AC^2) = sqrt(100 + 36) = 2sqrt(34), adding the two known lengths' squares instead of subtracting, as if AB were a leg of the right angle rather than the hypotenuse.
    • Why not D: This finds AB using only the horizontal coordinate difference, |5 - (-3)| = 8, forgetting to also include the vertical difference in the distance formula, then applies Pythagoras with this wrong value of AB.

  6. Question 6Answer: D

    1. In cyclic quadrilateral PQRS with vertices in order, opposite angles sum to 180 degrees: the opposite pairs are P and R, and Q and S.
    2. Using P and R: (3x + 10) + (x + 50) = 180, so 4x + 60 = 180, giving x = 30.
    3. Then angle Q = 2(30) - 5 = 55 degrees.
    4. Since Q and S are opposite, angle S = 180 - 55 = 125 degrees, option D.
    • Why not A: This correctly finds x = 30 and then angle Q = 55 degrees using the P and R equation, but stops there and reports angle Q instead of continuing to find angle S, which is what the question actually asks for.
    • Why not B: This sets angle P + angle R equal to 360 degrees, the angle sum of the whole quadrilateral, instead of 180 degrees, the rule that applies specifically to a pair of opposite angles in a cyclic quadrilateral, giving x = 75 and a wrong angle S of 35 degrees.
    • Why not C: This correctly finds x = 30, but then pairs S with P instead of with its true opposite angle Q, computing 180 - angle P = 180 - 100 = 80 instead of 180 - angle Q.

  7. Question 7Answer: A

    1. By the alternate segment theorem, the tangent-chord angle equals the angle in the alternate segment, so angle TRQ = 47 degrees.
    2. In triangle TQR, the three angles sum to 180 degrees: angle TRQ + angle RTQ + angle TQR = 180.
    3. So 47 + 71 + angle TQR = 180, which gives angle TQR = 180 - 118 = 62 degrees, option A.
    • Why not B: This uses the tangent-chord angle of 47 degrees directly as angle TQR, misapplying the alternate segment theorem to the wrong vertex of the triangle; the theorem gives angle TRQ, not angle TQR.
    • Why not C: This treats angle TQR as an exterior angle equal to the sum of the other two given angles, 71 + 47 = 118, instead of using the fact that the three angles of triangle TQR sum to 180 degrees.
    • Why not D: This uses only one of the two given angles, computing 180 - 71 = 109, and drops angle TRQ = 47 degrees entirely from the angle sum.

  8. Question 8Answer: C

    1. The distance from C to P is CP = sqrt((14-2)^2 + (4-(-1))^2) = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13.
    2. The tangent at the point of contact T is perpendicular to the radius CT, so triangle CTP is right-angled at T with hypotenuse CP.
    3. By Pythagoras, CT^2 + PT^2 = CP^2, so 25 + PT^2 = 169, giving PT^2 = 144.
    4. So the tangent length PT = 12, option C.
    • Why not A: This reports the distance CP = 13 itself as the tangent length, ignoring that the tangent meets the radius at a right angle at the point of contact and forms a separate side of a right-angled triangle.
    • Why not B: This computes the tangent length as CP - radius = 13 - 5 = 8, treating the centre, the point of contact and the external point as if they lay on a straight line, instead of applying Pythagoras to the right-angled triangle they actually form.
    • Why not D: This correctly finds PT^2 = 144 by Pythagoras but forgets to take the square root at the end, giving the squared length instead of the tangent length itself.

  9. Question 9Answer: D

    1. The line 4x + 3y = 5 has gradient -4/3, so the required line, being parallel to it, through (3, 4) also has gradient -4/3.
    2. Using y - 4 = -4/3(x - 3) and simplifying gives 4x + 3y = 24.
    3. Setting y = 0 gives the x-intercept A = (6, 0); setting x = 0 gives the y-intercept B = (0, 8).
    4. By Pythagoras, AB = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10, option D.
    • Why not A: This distributes -4/3 across (x - 3) as -4/3 x - 3, only multiplying the x-term by -4/3 and leaving the constant term unmultiplied, which gives the wrong line 4x + 3y = 3 and a much smaller AB.
    • Why not B: This distributes -4/3 across (x - 3) as -4/3 x + 3, correctly making the constant term positive but forgetting to multiply the 3 by 4/3, which gives the wrong line 4x + 3y = 21 instead of 4x + 3y = 24.
    • Why not C: This adds the two intercepts directly, 6 + 8 = 14, instead of using Pythagoras to find the straight-line distance between the two points A and B.

  10. Question 10Answer: C

    1. Since C and D lie on the same side of AB (the same segment), angle ACB = angle ADB: 4x - 15 = 2x + 25.
    2. Solving gives 2x = 40, so x = 20. Then angle ACB = 4(20) - 15 = 65 degrees.
    3. Angle AOB at the centre, subtended by the same arc as angle ACB, is twice the circumference angle: angle AOB = 2 x 65 = 130 degrees, option C.
    • Why not A: This correctly finds angle ACB = 65 degrees from the same-segment equality, but stops there and reports it instead of continuing to find angle AOB, which is what the question asks for.
    • Why not B: This correctly finds angle ACB = 65 degrees but then applies the cyclic-quadrilateral supplementary rule (180 - 65 = 115) instead of the correct rule that the angle at the centre is twice the angle at the circumference.
    • Why not D: This solves 4x - 15 = 2x + 25 correctly as far as 2x = 40, but then divides by 4 instead of 2, giving x = 10 instead of x = 20, which leads to a wrong angle ACB of 25 degrees and a wrong angle AOB of 50 degrees.

  11. Question 11Answer: B

    1. Complete the square: x^2 + 4x = (x+2)^2 - 4, and y^2 - 10y = (y-5)^2 - 25.
    2. Substituting back: (x+2)^2 - 4 + (y-5)^2 - 25 + k = 0, which rearranges to (x+2)^2 + (y-5)^2 = 29 - k.
    3. This must equal r^2 = 49 (since the radius is 7): 29 - k = 49.
    4. So k = 29 - 49 = -20, option B.
    • Why not A: This sets 29 - k equal to the radius, 7, instead of the radius squared, 49, forgetting that the completed-square form gives r^2 on the right-hand side, not r.
    • Why not C: This completes the square with the wrong sign, writing x^2 + 4x = (x+2)^2 + 4 and y^2 - 10y = (y-5)^2 + 25 (adding the compensating constants instead of subtracting them), which leads to -29 - k = 49 instead of 29 - k = 49.
    • Why not D: This correctly reaches 29 - k = 49, but then drops the negative sign when isolating k, giving k = 20 instead of k = -20.

  12. Question 12Answer: A

    1. Since OA and OB are both radii, triangle OAB is isosceles, so angle OBA = angle OAB = 35 degrees.
    2. The angle sum of triangle OAB gives angle AOB = 180 - 35 - 35 = 110 degrees.
    3. Since C lies on the major arc, angle ACB is subtended by the same (minor) arc as angle AOB, so angle ACB = angle AOB / 2 = 110 / 2 = 55 degrees, option A.
    • Why not B: This subtracts only one base angle from 180 (180 - 35 = 145) instead of both, forgetting that triangle OAB is isosceles with OA = OB, so it has two equal base angles of 35 degrees, not one; halving this wrong centre angle (145 / 2) still gives the wrong circumference angle of 72.5 degrees.
    • Why not C: This correctly uses the isosceles triangle to find angle AOB = 110 degrees, but then forgets that the angle at the centre is twice the angle at the circumference, reporting the centre angle itself instead of halving it.
    • Why not D: This doubles the given base angle OAB directly (35 x 2 = 70), as though it were already a circumference angle needing doubling to reach the centre, skipping the isosceles-triangle step that actually finds angle AOB.

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