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Home > Introduction to Pre-Calculus > Introduction to Graphing Functions > Examples of Circle and Semi-circle functions

Examples of Circle and Semi-circle functions

We look at a number of examples of circle and semi-circle functions, sketch their graphs, work out their domains and ranges, determine the centre and radius of a circle given its function, etc.

Here’s how you can test the circles and semi-circle functions

 

Example 1 : State the domain and range for each of the following functions and sketch its graph.

i). x2 + y2 = 25         Centre (0, 0),   r = 5

circle function example1

Domain: {x | −5 ≤ x ≤ 5}

Range: {y | −5 ≤ y ≤ 5}

 

ii) (x-2)2 + (y-3)2 = 25 = 52         Centre (2, 3),   r = 5

x ranges from -3 to 7

y ranges from -2 to 8

circle function example 2

Domain: {x | −3 ≤ x ≤ 7}

Range: {y | −2 ≤ x ≤ 8}

 

iii) y = sqrt{25~-~x^2}

r = 5. This is a semi-circle above x-axis.

x ranges from -5 to 5, but y ranges only in the positive direction, from 0 to 5.

semi-circle function example1

Domain: {x | −5 ≤ x ≤ 5}

Range: {y | 0 ≤ y ≤ 5}

iv). y = -~sqrt{64~-~x^2}

r = 8. This is a semi-circle below x-axis.

semi-circle function example2

Domain: {x | −8 ≤ x ≤ 8}

Range: {y | −8 ≤ y ≤ 0}

 

Example 2 : Write the equation for the circle with –

i) centre (0, 0) and radius 6.

(x-0)2 + (y-0)2 = 62

x2 + y2 = 36

 

ii) centre (-1, 5) and radius 9

(x+1)2 + (y-5)2 = 92

x2 + 2x + 1 + y2 -10y + 25 = 81

x2 + y+ 2x – 10y + 26 = 81

x2 + y+ 2x – 10y – 55 = 0  is the general equation of a circle

 

Example 3 : Given general equation of a circle is x2 + y+ 2x – 4y – 11 = 0, find the centre and the radius.

Complete square in general form to solve this equation –

x2  + 2x + 1 + y– 4y + 4 – 1 – 4 – 11 = 0

(x + 1)2 + (y – 2)2 – 1 – 4 – 11 = 0

(x + 1)2 + (y – 2)2 – 16 = 0

(x + 1)2 + (y – 2)2 = 16 = 42

So centre of this circle is (-1, 2) and the radius is 4.

 

Hint: We could also use the formula x2 + y+ 2fx + 2gy + c = 0, where centre is (-f, -g) and radius is sqrt{g^2~+~f^2~-~c} to solve the above circle equation x2 + y+ 2x – 4y – 11 = 0

Here 2f = 2, 2g = -4 and c = -11

So f = 1, g = -2, and centre is (-1, 2), and

radius = sqrt{(-2)^2~+~1^2~+~11}

radius = sqrt{16}  =  4