Mathematics · OCR H240 · Topic 1.03

Coordinate geometry in the x-y plane

Equations of straight lines and circles, and curves given by parametric equations.

  • 4 sub-topics
  • 2 in Year 12, 2 in Year 13
  • 83 practice questions
  • 4 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 1.03a-cStraight linesY12
  2. 1.03d-fCirclesY12
  3. 1.03gParametric equations of curvesY13
  4. 1.03hParametric equations in contextY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 83 questions in this topic, free, no account needed.

Coordinate geometry in the (x, y) plane

Easy

Equation of a line · Year 12 · 3 marks

  1. Find the equation of the line through (2,3)(2, 3) and (6,11)(6, 11) in the form y=mx+cy = mx + c.
    (3)
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  1. Gradient is change in yy over change in xx.m=11−36−2=2m = \dfrac{11 - 3}{6 - 2} = 2
  2. Use y−y1=m(x−x1)y - y_1 = m(x - x_1).y−3=2(x−2)y - 3 = 2(x - 2)
  3. y=2x−1y = 2x - 1

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Coordinate geometry in the (x, y) plane

Medium

Tangent to a circle · Year 12 · 4 marks

The circle CC has equation (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25.
  1. Find the equation of the tangent to CC at (6,2)(6, 2).
    (4)
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  1. The radius goes from the centre (3,−2)(3, -2) to the point.Radius gradient =2−(−2)6−3= \dfrac{2 - (-2)}{6 - 3}=43{}= \dfrac{4}{3}
  2. A tangent is perpendicular to the radius.Tangent gradient =−34= -\tfrac{3}{4}
  3. y−2=−34(x−6)y - 2 = -\tfrac{3}{4}(x - 6)
  4. 3x+4y−26=03x + 4y - 26 = 0

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Coordinate geometry in the (x, y) plane

Hard

A circle through three points · Year 12 · 6 marks

A(0,2)A(0, 2), B(4,6)B(4, 6) and C(4,2)C(4, 2)
  1. (a)
    Show that angle ACBACB is a right angle.
    (2)
  2. (b)
    Hence find the equation of the circle through AA, BB and CC.
    (4)
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Part (a)2 marks

  1. Look at the gradients: 0 and undefined.ACAC is horizontal (y=2y = 2) and CBCB is vertical (x=4x = 4)
  2. So ACAC and CBCB are perpendicular

Part (b)4 marks

  1. The angle in a semicircle is a right angle.ABAB is a diameter
  2. Centre = midpoint of AB=(2,4)AB = (2, 4)
  3. Distance from the centre to AA.r2=(2−0)2+(4−2)2=8r^2 = (2 - 0)^2 + (4 - 2)^2 = 8
  4. (x−2)2+(y−4)2=8(x - 2)^2 + (y - 4)^2 = 8

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