Seneca
Sign up

Finding the Centre & Radius

Maths: Edexcel A Level Pure Maths, topic 3.2.2

How do you find the Centre and Radius of a Circle?

To find the centre and radius of a circle from its expanded equation, we need to complete the square.

What is the expanded equation of a circle?

What is the expanded equation of a circle?

  • If we expand the general equation of a circle with centre (a,b)(a,b) and radius rr we get:
  • (x−a)2+(y−b)2=r2(x - a)^2 + (y-b)^2 = r^2
  • x2−2ax+a2+y2−2by+b2=r2x^2 -2ax + a^2 +y^2 -2by +b^2=r^2
  • x2+y2−2ax−2by+a2+b2−r2=0x^2 +y^2 -2ax -2by +a^2+b^2-r^2 = 0
What is the expanded equation of a circle?

What is the expanded equation of a circle?

  • If we make the substitutions:
  • a=−fa = -f
  • b=−gb = -g
  • a2+b2−r2=ca^2+b^2-r^2= c
  • Then the expanded form of the equation of a circle is given by:
  • x2+y2+2fx+2gy+c=0x^2+y^2 +2fx +2gy+c = 0
How do you find the centre and radius of a circle?

How do you find the centre and radius of a circle?

  • If the equation of a circle is in the expanded form, we can complete the square and find the circle in the general form:
  • (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2 = r^2.
  • The centre of the circle is at the point (a,b)(a,b) and the radius is of length rr.
Example

Example

  • What is the centre and radius of the circle with equation x2+y2−4x−6y+4=0x^2 + y^2-4x-6y+4 = 0?
Group together $$x$$ and $$y$$ terms

Group together xx and yy terms

  • To begin, we group together the terms containing xx and the terms containing yy:
  • x2−4x+y2−6y+4=0x^2 -4x + y^2 - 6y + 4 = 0
Complete the square for terms containing $$x$$

Complete the square for terms containing xx

  • We then complete the square for the xx terms alone:
  • x2−4x=(x−2)2−4x^2 -4x = (x-2)^2 -4
Complete the square for terms containing $$7$$

Complete the square for terms containing 77

  • We then complete the square for the yy terms alone:
  • y2−6y=(y−3)2−9y^2 -6y = (y-3)^2 -9
Substitute in

Substitute in

  • We then substitute these back into the equation:
  • (x−2)2−4+(y−3)2−9+4=0(x-2)^2 - 4 + (y-3)^2 -9 +4=0
Simplify

Simplify

  • Simplifying the constants gives:
  • (x−2)2+(y−3)2−9=0(x-2)^2 + (y-3)^2 -9 = 0
  • Then adding 9 (which is 32) to each side, we get the equation of the circle in its general form:
  • (x−2)2+(y−3)2=32(x-2)^2 + (y-3)^2 = 3^2
Inspect

Inspect

  • Comparing this to the general form of the equation, we can see that the centre of the circle is at the point (2,3) and the radius is equal to 3.

Test yourself on Finding the Centre & Radius

  • What is the xx-coordinate of the centre of the circle with equation x2+y2−10y=0x^2 + y^2 -10y = 0?

Practise this topic for free