3.2.2

Finding the Centre & Radius

Test yourself

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.

Illustrative background for What is the expanded equation of a circle?Illustrative background for What is the expanded equation of a circle? ?? "content

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:
    • (xa)2+(yb)2=r2(x - a)^2 + (y-b)^2 = r^2
    • x22ax+a2+y22by+b2=r2x^2 -2ax + a^2 +y^2 -2by +b^2=r^2
    • x2+y22ax2by+a2+b2r2=0x^2 +y^2 -2ax -2by +a^2+b^2-r^2 = 0
Illustrative background for What is the expanded equation of a circle?Illustrative background for What is the expanded equation of a circle? ?? "content

What is the expanded equation of a circle?

  • If we make the substitutions:
    • a=fa = -f
    • b=gb = -g
    • a2+b2r2=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
Illustrative background for How do you find the centre and radius of a circle?Illustrative background for How do you find the centre and radius of a circle? ?? "content

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:
    • (xa)2+(yb)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.
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Example

  • What is the centre and radius of the circle with equation x2+y24x6y+4=0x^2 + y^2-4x-6y+4 = 0?
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$$y$$ termsIllustrative background for Group together $$x$$ and 
$$y$$ terms ?? "content

Group together xx and yy terms

  • To begin, we group together the terms containing xx and the terms containing yy:
    • x24x+y26y+4=0x^2 -4x + y^2 - 6y + 4 = 0
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Complete the square for terms containing xx

  • We then complete the square for the xx terms alone:
    • x24x=(x2)24x^2 -4x = (x-2)^2 -4
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Complete the square for terms containing 77

  • We then complete the square for the yy terms alone:
    • y26y=(y3)29y^2 -6y = (y-3)^2 -9
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Substitute in

  • We then substitute these back into the equation:
    • (x2)24+(y3)29+4=0(x-2)^2 - 4 + (y-3)^2 -9 +4=0
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Simplify

  • Simplifying the constants gives:
    • (x2)2+(y3)29=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:
    • (x2)2+(y3)2=32(x-2)^2 + (y-3)^2 = 3^2
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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.

Jump to other topics

1Proof

2Algebra & Functions

2.1Powers & Roots

2.2Quadratic Equations

2.3Inequalities

2.4Polynomials

2.5Graphs

2.6Functions

2.7Transformation of Graphs

2.8Partial Fractions (A2 Only)

3Coordinate Geometry

4Sequences & Series

5Trigonometry

6Exponentials & Logarithms

7Differentiation

8Integration

9Numerical Methods

10Vectors

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