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General Equation of a Circle0%

Circles · Topic 4 of 7

General Equation of a Circle

Video lesson · from 11:512 worked examples

One lesson video covers all of Circles, so it opens at 11:51 for this topic — not from the beginning.

Theory

The equation of a circle can be written in expanded form.

The circle with centre (-2, 3) and radius 5 units has equation:

(x+2)2+(y−3)2=25(x+2)^2+(y-3)^2=25
x2+4x+4+y2−6y+9=25x^2+4x+4+y^2-6y+9=25
x2+4x+y2−6y=12x^2+4x+y^2-6y=12
x2+y2+4x−6y−12=0x^2+y^2+4x-6y-12=0

This form of the equation is called the general equation of a circle.

Working in reverse from the general equation you can find the centre and radius using completing the square.

x2+4x+y2−6y=12x^2+4x+y^2-6y=12
(x+2)2−4+(y−3)2−9=12(x+2)^2-4+(y-3)^2-9=12
(x+2)2+(y−3)2=25(x+2)^2+(y-3)^2=25

Centre (-2, 3), Radius = 5 units

Given what we have learned about the general equation of a circle so far, we can use this as a quicker way to find the centre and radius.

x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0
x2+y2+2gx+2fy=−cx^2+y^2+2gx+2fy=-c
(x+g)2−g2+(y+f)2−f2=−c(x+g)^2-g^2+(y+f)^2-f^2=-c
(x+g)2+(y+f)2=g2+f2−c(x+g)^2+(y+f)^2=g^2+f^2-c

This is now in the form (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2, giving:

  • Centre: (−g,−f)(-g, -f)
  • Radius: r=g2+f2−cr = \sqrt{g^2+f^2-c}
  • Valid if: g2+f2−c>0g^2+f^2-c > 0

⚠️ Common Examiner Traps

  • Sign of the centre: the centre is (−g,−f)(-g,-f), so the coefficients 2g2g and 2f2f flip sign twice. For x2+y2+4x−8y+7=0x^2+y^2+4x-8y+7=0, 2g=42g=4 gives g=2g=2 and the centre has x=−2x = -2.
  • Halve before you negate: reading the centre straight off as (−4,8)(-4, 8) instead of (−2,4)(-2, 4) is the classic slip.
  • Show your working: stating the centre and radius with no supporting working will not gain full marks. State gg, ff and cc, then the centre and radius.
  • Check it is a circle: if g2+f2−c≤0g^2+f^2-c \leq 0 there is no circle. When asked to explain why an equation is not a circle, work the value out and say it is negative.
  • Use earlier parts: a very common failure is not carrying an earlier answer forward — most candidates who lose marks here never use their part (a) result to find the centre. If you have just found a midpoint or an equation, it is almost certainly needed next.

Worked examples

Example 1

Find the radius and centre of the circle with equation x2+y2+4x−8y+7=0x^2+y^2+4x-8y+7=0.

Compare with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0:

  • 2g=4  ⟹  g=22g = 4 \implies g = 2
  • 2f=−8  ⟹  f=−42f = -8 \implies f = -4
  • c=7c = 7

Centre: (−g,−f)=(−2,4)(-g, -f) = (-2, 4)

Radius: r=g2+f2−c=22+(−4)2−7=4+16−7=13r = \sqrt{g^2+f^2-c} = \sqrt{2^2 + (-4)^2 - 7} = \sqrt{4 + 16 - 7} = \sqrt{13} units

Example 2

Explain why x2+y2+4x−8y+29=0x^2+y^2+4x-8y+29=0 is not the equation of a circle.

Compare with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0:

  • 2g=4  ⟹  g=22g = 4 \implies g = 2
  • 2f=−8  ⟹  f=−42f = -8 \implies f = -4
  • c=29c = 29

Check the radius condition g2+f2−c>0g^2+f^2-c > 0:

g2+f2−c=22+(−4)2−29=4+16−29=20−29=−9g^2+f^2-c = 2^2 + (-4)^2 - 29 = 4 + 16 - 29 = 20 - 29 = -9

Since g2+f2−c<0g^2+f^2-c < 0, we would be taking the square root of a negative number for the radius. Therefore, it does not represent a valid circle.