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CIE A-Level Maths Study Notes

1.3.4 Equation of a Circle

The equation of a circle is a fundamental concept in coordinate geometry, essential for solving various geometrical problems. We will explore the standard and general forms of a circle's equation, focusing on finding the centre and radius and converting between these forms.

Standard Form of a Circle's Equation

  • Equation: (xa)2+(yb)2=r2(x-a)^2 + (y-b)^2 = r^2
  • Centre: (a,b)(a, b)
  • Radius: rr

In the standard form, aa and bb represent the coordinates of the centre of the circle, and rr is the radius.

General Form of a Circle's Equation

  • Equation: x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0
  • Centre:(g,f) \left(-g, -f\right)
  • Radius: g2+f2c\sqrt{g^2 + f^2 - c}

Key Concepts

  • Completing the Square: A method to convert the general form of a circle's equation to the standard form.
  • Tangents and Radius: Tangents to a circle are always perpendicular to the radius at the point of contact.
  • Right-Angled Triangle in Circle: If a right-angled triangle is inscribed in a circle, its hypotenuse is the diameter of the circle.

Example 1

Given the equation of a circle x2+y2+6x8y+9=0x^2 + y^2 + 6x - 8y + 9 = 0, find the centre and radius.

Solution:

1. Convert to standard form:

x2+6x+y28y=9x^2 + 6x + y^2 - 8y = -9

(x+3)29+(y4)216=9(x + 3)^2 - 9 + (y - 4)^2 - 16 = -9

(x+3)2+(y4)2=16(x + 3)^2 + (y - 4)^2 = 16

2. Centre: (3,4)(-3, 4)

3. Radius: 16=4\sqrt{16} = 4

graph of circle


Example 2

Find the centre and radius of the circle given by the equation x2+y212x+4y+36=0x^2 + y^2 - 12x + 4y + 36 = 0.

Solution:

1. Convert to standard form:

x212x+y2+4y=36x^2 - 12x + y^2 + 4y = -36

(x6)236+(y+2)24=36(x - 6)^2 - 36 + (y + 2)^2 - 4 = -36

(x6)2+(y+2)2=4(x - 6)^2 + (y + 2)^2 = 4

2. Centre: (6,2)(6,-2)

3. Radius: 4=2\sqrt{4} = 2

graph of a circle

Example 3

Task: A circle has the equation x2+y24x+6y12=0x^2 + y^2 - 4x + 6y - 12 = 0. Determine its centre and radius.

Solution:

1. Rearrange to standard form:

x24x+y2+6y=12x^2 - 4x + y^2 + 6y = 12

(x2)24+(y+3)29=12(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12

(x2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25

2. Centre: (2,3)(2,-3)

3. Radius: 25=5\sqrt{25} = 5

graph of a circle
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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