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TMUA Coordinate geometry questions

9 original TMUA coordinate geometry questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Tangent means distance from centre = radiusHow two circles meetPerpendicular bisector: equidistant from both points

Question 1TMUA Paper 1Standard

The line y=x+cy = x + c is a tangent to the circle x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0. What is the sum of all possible values of cc?
  1. A−10-10
  2. B−5-5
  3. C00
  4. D55
  5. E1010
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Answer: A · Method: Tangent means distance from centre = radius

  1. Complete the square: (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25, so the centre is (3,−2)(3, -2) and the radius is 55.
  2. The distance from the centre to x−y+c=0x - y + c = 0 is ∣3+2+c∣2\dfrac{\lvert 3 + 2 + c \rvert}{\sqrt{2}}. Setting this equal to 55 gives c=−5±52c = -5 \pm 5\sqrt{2}.
  3. The sum of the two values is −10-10.

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Question 2TMUA Paper 1Standard

Circle C1C_1 has equation (x−1)2+(y+2)2=9(x - 1)^2 + (y + 2)^2 = 9. Circle C2C_2 has centre (7,6)(7, 6) and radius rr, where rr is a positive integer. For how many values of rr do the two circles meet at exactly two points?
  1. A44
  2. B55
  3. C66
  4. D77
  5. E1111
  6. F1313
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Answer: B · Method: How two circles meet

  1. The centres are (1,−2)(1, -2) and (7,6)(7, 6), a distance 62+82=10\sqrt{6^2 + 8^2} = 10 apart. C1C_1 has radius 33.
  2. Two circles meet at two points exactly when the distance between centres is strictly between the difference and the sum of the radii: ∣r−3∣<10<r+3\lvert r - 3 \rvert < 10 < r + 3.
  3. This gives 7<r<137 < r < 13, so r=8,9,10,11,12r = 8, 9, 10, 11, 12: five values.

Why the tempting wrong answers are wrong

  • D: At r=7r = 7 the circles touch externally and at r=13r = 13 internally: one point each, not two.

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Question 3TMUA Paper 1Standard

The line y=mxy = mx is a tangent to the circle (x−4)2+(y−2)2=4(x - 4)^2 + (y - 2)^2 = 4. What is the sum of all possible values of mm?
  1. A43\dfrac{4}{3}
  2. B00
  3. C34\dfrac{3}{4}
  4. D22
  5. E83\dfrac{8}{3}
  6. F−43-\dfrac{4}{3}
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Answer: A · Method: Tangent means distance from centre = radius

  1. The centre is (4,2)(4, 2) and the radius is 22. The line is tangent exactly when its distance from the centre equals the radius.
  2. Distance from (4,2)(4, 2) to mx−y=0mx - y = 0 is ∣4m−2∣m2+1=2\dfrac{\lvert 4m - 2 \rvert}{\sqrt{m^2 + 1}} = 2, so (4m−2)2=4(m2+1)(4m - 2)^2 = 4(m^2 + 1).
  3. This simplifies to 12m2−16m=012m^2 - 16m = 0, so m=0m = 0 or m=43m = \tfrac{4}{3}. The sum is 43\tfrac{4}{3}.

Why the tempting wrong answers are wrong

  • B: m=0m = 0 is one tangent (the line y=0y = 0 touches the circle's bottom), but there is a second one too.

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Question 4TMUA Paper 1Standard

Circle C1C_1 has equation x2+y2+4x−4y−8=0x^2 + y^2 + 4x - 4y - 8 = 0. Circle C2C_2 has centre (−10,8)(-10, 8) and radius rr. There are two values of rr for which the circles touch each other. What is the product of these two values?
  1. A4040
  2. B8383
  3. C8484
  4. D8585
  5. E100100
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Answer: C · Method: How two circles meet

  1. C1C_1 has centre (−2,2)(-2, 2) and radius 44. The distance between the centres is 1010.
  2. They touch externally when r+4=10r + 4 = 10, so r=6r = 6, and internally when r−4=10r - 4 = 10, so r=14r = 14.
  3. The product is 8484.

Why the tempting wrong answers are wrong

  • E: One value is when they touch externally (r+r1=dr + r_1 = d) and the other internally (r−r1=dr - r_1 = d).

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Question 5TMUA Paper 1Standard

Circle C1C_1 has equation x2+y2−6x+2y−26=0x^2 + y^2 - 6x + 2y - 26 = 0. Circle C2C_2 has centre (−9,−6)(-9, -6) and radius rr, where rr is a positive integer. For how many values of rr do the circles meet at exactly two points?
  1. A66
  2. B1010
  3. C1111
  4. D1212
  5. E1313
Show the answer and worked solution

Answer: C · Method: How two circles meet

  1. C1C_1 has centre (3,−1)(3, -1) and radius 66. The centres are 144+25=13\sqrt{144 + 25} = 13 apart.
  2. Two circles meet at two points exactly when the distance between centres is strictly between the difference and the sum of the radii: ∣r−6∣<13<r+6\lvert r - 6 \rvert < 13 < r + 6.
  3. So 7<r<197 < r < 19: that's 1111 integer values.

Why the tempting wrong answers are wrong

  • E: At r=7r = 7 and r=19r = 19 the circles only touch, at one point.

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Question 6TMUA Paper 1Standard

Circle C1C_1 has equation (x+4)2+(y−4)2=16(x + 4)^2 + (y - 4)^2 = 16. Circle C2C_2 has centre (8,9)(8, 9) and radius rr. There are two values of rr for which the circles touch each other. What is the product of these two values?
  1. A5252
  2. B152152
  3. C153153
  4. D154154
  5. E169169
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Answer: C · Method: How two circles meet

  1. C1C_1 has centre (−4,4)(-4, 4) and radius 44. The distance between the centres is 1313.
  2. They touch externally when r+4=13r + 4 = 13, so r=9r = 9, and internally when r−4=13r - 4 = 13, so r=17r = 17.
  3. The product is 153153.

Why the tempting wrong answers are wrong

  • E: One value is when they touch externally (r+r1=dr + r_1 = d) and the other internally (r−r1=dr - r_1 = d).

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Question 7TMUA Paper 1Standard

Circle C1C_1 has equation (x−2)2+(y−2)2=9(x - 2)^2 + (y - 2)^2 = 9. Circle C2C_2 has centre (10,17)(10, 17) and radius rr, where rr is a positive integer. For how many values of rr do the circles meet at exactly two points?
  1. A33
  2. B44
  3. C55
  4. D66
  5. E77
Show the answer and worked solution

Answer: C · Method: How two circles meet

  1. C1C_1 has centre (2,2)(2, 2) and radius 33. The centres are 64+225=17\sqrt{64 + 225} = 17 apart.
  2. Two circles meet at two points exactly when the distance between centres is strictly between the difference and the sum of the radii: ∣r−3∣<17<r+3\lvert r - 3 \rvert < 17 < r + 3.
  3. So 14<r<2014 < r < 20: that's 55 integer values.

Why the tempting wrong answers are wrong

  • E: At r=14r = 14 and r=20r = 20 the circles only touch, at one point.

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Question 8TMUA Paper 1Standard

Circle C1C_1 has equation x2+y2−4x−6y+4=0x^2 + y^2 - 4x - 6y + 4 = 0. Circle C2C_2 has centre (−1,−1)(-1, -1) and radius rr, where rr is a positive integer. For how many values of rr do the circles meet at exactly two points?
  1. A33
  2. B44
  3. C55
  4. D66
  5. E77
Show the answer and worked solution

Answer: C · Method: How two circles meet

  1. C1C_1 has centre (2,3)(2, 3) and radius 33. The centres are 9+16=5\sqrt{9 + 16} = 5 apart.
  2. Two circles meet at two points exactly when the distance between centres is strictly between the difference and the sum of the radii: ∣r−3∣<5<r+3\lvert r - 3 \rvert < 5 < r + 3.
  3. So 2<r<82 < r < 8: that's 55 integer values.

Why the tempting wrong answers are wrong

  • E: At r=2r = 2 and r=8r = 8 the circles only touch, at one point.

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Question 9TMUA Paper 1Standard

The perpendicular bisector of the line segment joining A(−2,3)A(-2, 3) and B(−4,−4)B(-4, -4) crosses the yy-axis at (0,k)(0, k). What is kk?
  1. A−1914-\dfrac{19}{14}
  2. B−12-\dfrac{1}{2}
  3. C−514-\dfrac{5}{14}
  4. D1914\dfrac{19}{14}
  5. E1010
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Answer: A · Method: Perpendicular bisector: equidistant from both points

  1. The midpoint of ABAB is (−3,−12)(-3, -\dfrac{1}{2}) and ABAB has gradient 72\dfrac{7}{2}, so the bisector has gradient −27-\dfrac{2}{7}.
  2. Alternatively: (0,k)(0, k) is the same distance from AA and BB.
  3. 4+(k−3)2=16+(k+4)24 + (k - 3)^2 = 16 + (k + 4)^2 gives k=−1914k = -\dfrac{19}{14}.

Why the tempting wrong answers are wrong

  • B: That's the yy-coordinate of the midpoint, which isn't on the yy-axis.
  • E: The bisector is perpendicular to ABAB, so its gradient is the negative reciprocal of ABAB's.

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