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Topics · Equation of a Circle

Equations of circles in DSE Maths Paper 2

This topic puts circles on the coordinate plane. You need to write the equation of a circle from its centre and radius or from other conditions, read the centre and radius off the general form, and decide how a circle and a straight line meet.

What this topic covers

  • The equation of a circle from its centre and radius
  • Centre and radius from the general form x² + y² + Dx + Ey + F = 0
  • A circle through given points, or with its centre on a given line
  • Intersections of a circle and a straight line; tangents

Two sample questions

Sample 1

P( negative 5, negative 17) and Q(5, 7) are the two endpoints of a diameter of a circle. The equation of the circle is

  1. Ax squared plus y squared plus 10x plus 194 equals 0.
  2. Bx squared plus y squared plus 10x minus 144 equals 0.
  3. Cx squared plus y squared plus 10y minus 144 equals 0.
  4. Dx squared plus y squared plus 10y plus 194 equals 0.
Show the answer and solution

Answer: C

Worked solution

Centre=\displaystyle =mid-point of PQ\displaystyle PQ=(−5+52,−17+72)\displaystyle {}= \left(\dfrac{-5+5}{2}, \dfrac{-17+7}{2}\right)=(0,−5)\displaystyle {}= (0, -5)Radius=12PQ\displaystyle {}= \dfrac{1}{2}PQ=12(5−(−5))2+(7−(−17))2\displaystyle {}= \dfrac{1}{2}\sqrt{(5-(-5))^2+(7-(-17))^2}=13\displaystyle {}= 13(x−0)2+(y−(−5))2=132\displaystyle (x - 0)^2 + (y - (-5))^2 = 13^2⇒x2+y2+10y−144=0\displaystyle \Rightarrow x^2 + y^2 + 10y - 144 = 0

Sample 2

A circle passes through the points A( negative 2, 4) and B( negative 4, 6), and its centre lies on the line x plus y equals 0. The equation of the circle is

  1. Ax squared plus y squared minus 8x plus 8y plus 36 equals 0.
  2. Bx squared plus y squared plus 8x minus 8y plus 28 equals 0.
  3. Cx squared plus y squared plus 8x minus 8y plus 36 equals 0.
  4. Dx squared plus y squared minus 8x plus 8y plus 28 equals 0.
Show the answer and solution

Answer: B

Worked solution

At A: 20−2D+4E+F=0\displaystyle A: \ 20 - 2D + 4E + F = 0 \quadat B: 52−4D+6E+F=0\displaystyle B: \ 52 - 4D + 6E + F = 0Subtracting eliminates F: 2(D−E)=32⇒D−E=16\displaystyle F: \ 2(D - E) = 32 \Rightarrow D - E = 16Centre (−D2,−E2)\displaystyle \left(-\dfrac{D}{2}, -\dfrac{E}{2}\right) on x+y=0: D+E=0⇒D=8, E=−8\displaystyle x+y=0: \ D + E = 0 \Rightarrow D = 8, \ E = -8Back into B: F=−52−8(−4)−(−8)(6)=28\displaystyle B: \ F = -52 - 8(-4) - (-8)(6) = 28⇒x2+y2+8x−8y+28=0\displaystyle \Rightarrow x^2 + y^2 + 8x - 8y + 28 = 0

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Original material by Dsetopia, written from the EDB syllabus learning objectives. Not affiliated with the HKEAA; no exam content is reproduced.