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Equation of a circle

Equation of a circle

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Summary


Equation of a circle 

In a nutshell

The equation of a circle uses Pythagoras' theorem to relate any point (x,y)(x,y)(x,y)​ on the circumference of the circle to the radius of the circle. From the equation of a circle the centre and the radius of the circle can be determined. 

​

​

The equation of a circle with centre (0,0)(0,0)(0,0)​

When a circle has centre (0,0)(0,0)(0,0) and radius rrr, the equation of the circle is 

x2+y2=r2\boxed{{x^2} + {y^2} = {r^2}}x2+y2=r2​

​

Example 1 

The circumference of a circle with centre (0,0)(0,0)(0,0) passes through the point (3,4)(3,4)(3,4). What is the radius of the circle? 


​Draw a sketch using all the information given in the question.

Maths; Circles; KS5 Year 12; Equation of a circle


Using the equation x2+y2=r2{{x^2} + {y^2} = {r^2}}x2+y2=r2​, solve for rrr.

x2+y2=r232+42=r225=r2\begin{aligned}x^2 + y^2 &= r^2 \\3^2 + 4^2 &= r^2 \\25 &=r^2 \\\end {aligned}x2+y232+4225​=r2=r2=r2​​​

​r=5‾\underline{r=5}r=5​​​

​​​

​​​

The equation of a circle with centre (a,b)(a,b)(a,b)​

When a circle has centre (a,b)(a,b)(a,b)​ and radius rr r, the equation of the circle is ​

​(x−a)2+(y−b)2=r2\boxed{{(x-a)^2} + {(y-b)^2} = {r^2}}(x−a)2+(y−b)2=r2​


Example 2 

A circle has the equation (x−5)2+(y+6)2=90{{(x-5)^2} + {(y+6)^2} = {90}}(x−5)2+(y+6)2=90​. Write down the radius and centre of the circle, then show that the circle passes through (2,3)(2,3)(2,3).


The centre of the circle is (5,−6)‾\underline{(5,-6)}(5,−6)​.

The radius is 90=9×10=9×10=310‾\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = \underline{3\sqrt{10}}90​=9×10​=9​×10​=310​​

​

Now substitute the coordinates (2,3)(2,3)(2,3) into the equation of the circle to show that the circle passes through that point. 


(2−5)2+(3+6)2=90{{(2-5)^2} + {(3+6)^2} = {90}}(2−5)2+(3+6)2=90

(−3)2+(9)2=90{{(-3)^2} + {(9)^2} = {90}}(−3)2+(9)2=90

9+81=909 + 81 = {90}9+81=90


Therefore, the point does lie on the circle.


Maths; Circles; KS5 Year 12; Equation of a circle



Expanded form of the equation of a circle


You may also see the equation of a circle in its expanded form. 

​x2+y2+2fx+2gy+c=0{{x^2} + {y^2} + 2fx + 2gy +c= 0}x2+y2+2fx+2gy+c=0


This circle has centre (−f,−g)(-f,-g)(−f,−g) and radius f2+g2−c\sqrt{{f^2} + {g^2} - c}f2+g2−c​. To find the centre and radius of a circle with its equation in its expanded form, complete the square for the xxx and yyy terms. 


Example 3 

A circle has the equation x2+y2+6x−14y−17=0{{x^2} +{y^2} + 6x -14y -17 = 0}x2+y2+6x−14y−17=0. Find the centre and the radius of the circle. 


First the circle needs to be rearrange into the form (x−a)2+(y−b)2=r2{{(x-a)^2} + {(y-b)^2} = {r^2}}(x−a)2+(y−b)2=r2.​

​Rearrange the equation to collect xxx and yyy terms. ​

x2+6x+y2−14y=17{{x^2} + 6x + {y^2} -14y = 17}x2+6x+y2−14y=17​

Complete the square for xxx​ and yyy​ terms.​

(x+3)2−9+(y−7)2−49=17{({x+3)^2} -9 + {(y-7)^2} -49 = 17}(x+3)2−9+(y−7)2−49=17​

Collect all number terms to the right-hand side of the equation.

(x+3)2+(y−7)2=75{({x+3)^2} + {(y-7)^2} = 75}(x+3)2+(y−7)2=75​


The centre is (−3,7)‾\underline{(-3,7)}(−3,7)​​​​​

The radius is 75=25×3=25×3=53‾\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = \underline{5\sqrt{3}}75​=25×3​=25​×3​=53​​​


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FAQs - Frequently Asked Questions

If the equation of a circle is in its expanded form x^2 + y^2 + 2fx + 2gy +c= 0, how do you find the centre and radius of the circle?

If the equation of a circle is in its expanded form x^2 + y^2 + 2fx + 2gy +c= 0, complete the square for the x and y terms to find the centre and radius of the circle.

What is the equation of the circle with centre (a,b)?

The equation of a circle with centre (a,b) is (x-a)^2 + (y-b)^2 = r^2.

What is the equation of the circle with centre (0,0)?

The equation of a circle with centre (0,0) is x^2 + y^2 = r^2.

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