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11 Nonlinear relationships
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Stage 5.1-2

11.03 Graphs of circles

Lesson
Worksheet
Practice
Lesson

Transformations of circles

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Graphs of equations of the form \\ \left(x-h \right)^{2} + \left(y-k \right)^{2} = r^{2} (where h,\, k, and r are any number and r \neq 0) are called circles.

The circle defined by x^{2} + y^{2} = 1. It has a centre at (0,0) and a radius of 1 unit.

A circle can be vertically translated by increasing or decreasing the y-values by a constant number. However, the y-value together with the translation must be squared together. So to translate \\ x^{2} + y^{2} = 1 up by k units gives us x^{2} + \left(y-k \right)^{2} = 1.

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This graph shows x^{2} + y^{2} = 1 translated vertically up by 2 to get x^{2} + \left(y-2 \right)^{2} = 1, and down by 2 to get x^{2} + \left(y+2 \right)^{2} = 1.

Exploration

In the applet below, move the slider for k to vertically translate the circle and move the slider for r to adjust the radius.

Loading interactive...

When we subtract k from y then the circle is translated up k units. When we add k to y then the circle is translated down k units. As the radius increases the size of the circle also increases.

A circle can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value together with the translation must be squared together. That is, to translate \\ x^{2} + y^{2} = 1 to the left by h units we get \left(x+h \right)^{2}+ y^{2} = 1.

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This graph shows x^{2} + y^{2} = 1 translated horizontally left by 2 to get \left(x+2 \right)^{2} + y^{2}=1, and right by 2 to get \left(x-2 \right)^{2} + y^{2} = 1.

Notice that the centre of the circle x^{2} + y^{2} = 1 is at (0,0). Translating the circle will also translate the centre by the same amount. So the centre of \left(x-h \right)^{2} + \left(y-k \right)^{2} = r^{2} is at (h,k).

A circle can be scaled both vertically and horizontally by changing the value of r. In fact, r is the radius of the circle.

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This graph shows x^{2} + y^{2} = 1 expanded by a scale factor of 2 to get x^{2} + y^{2} = 4, and compressed by 2 to get x^{2} + y^{2} = \dfrac{1}{4}.

Examples

Example 1

Consider the circle with equation \left(x - 0.4\right)^{2} + \left(y + 3.8\right)^{2} = 2.

a

What is the centre of the circle?

Worked Solution
Create a strategy

Equations of the form (x-h)^{2} + (y-k)^{2} = r^{2} are circles with centre at (h,k).

Apply the idea

The equation \left(x - 0.4\right)^{2} + \left(y + 3.8\right)^{2} = 2 can be written as \left(x - 0.4\right)^{2} + \left(y - (-3.8)\right)^{2} = 2 and it is now of the form (x-h)^{2} + (y-k)^{2} = r^{2}. So we have h=0.4 and k=-3.8.

The centre of the circle is (0.4,-3.8).

b

What is the exact radius of the circle?

Worked Solution
Create a strategy

Equations of the form (x-h)^{2} + (y-k)^{2} = r^{2} are circles with radius of r.

Apply the idea

In part (a) the equation \left(x - 0.4\right)^{2} + \left(y + 3.8\right)^{2} = 2 can be written in the form (x-h)^{2} + (y-k)^{2} = r^{2}, we can say that:

\displaystyle r^{2}\displaystyle =\displaystyle 2
\displaystyle \sqrt{r^{2}}\displaystyle =\displaystyle \sqrt{2}Square root both sides
\displaystyle r\displaystyle =\displaystyle \sqrt{2}Evaluate

The radius of the circle is \sqrt{2} units.

Example 2

A circle has its centre at \left(3, - 2 \right) and a radius of 4 units.

a

Plot the graph for the given circle.

Worked Solution
Create a strategy

Plot the centre of the circle.

Count 4 units in each direction from the centre: left, right, up, and down and draw a round curve passing through each these points.

Apply the idea
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The centre of the circle is at the (3,-2). So plot the point (0,0).

Now we count 4 units in all four directions to get the points (3,2),\, (7,-2),\, (3,-6) and (-1,-2).

Connect the points using a round curve passing through the 4 points to graph the circle,

b

Write the equation of the circle in general form, x^{2} + y^{2} + ax +bx +c = 0.

Worked Solution
Create a strategy

Use the form \left(x - h\right)^{2} + \left(y - k\right)^{2} = r^2 and expand each perfect square.

Apply the idea

We are given that the centre of the circle is (3,-2) and the radius is 4. So h=3, \, k=-2 and \\ r=4.

\displaystyle \left(x - h\right)^{2} + \left(y - k\right)^{2}\displaystyle =\displaystyle r^{2}
\displaystyle \left(x - 3\right)^{2} + \left(y - (-2)\right)^{2}\displaystyle =\displaystyle 4^{2}Substitute h, \, k, and r
\displaystyle \left(x - 3\right)^{2} + \left(y +2\right)^{2}\displaystyle =\displaystyle 16Simplify
\displaystyle x^2 -6x + 9 + y^2 + 4y + 4\displaystyle =\displaystyle 16Expand the perfect squares
\displaystyle x^2 -6x + y^2 + 4y + 13\displaystyle =\displaystyle 16Combine like terms
\displaystyle x^2 -6x + y^2 + 4y + 13 -16\displaystyle =\displaystyle 16 -16Subtract 16 from both sides
\displaystyle x^2 -6x + y^2 + 4y -3\displaystyle =\displaystyle 0Evaluate
\displaystyle x^2 + y^2 -6x +4y -3\displaystyle =\displaystyle 0Write in general form

Example 3

Consider the equation of a circle given by x^{2} + 4 x + y^{2} + 6 y - 3 = 0.

a

Rewrite the equation of the circle in the form \left(x - h\right)^{2} + \left(y - k\right)^{2} = r^2.

Worked Solution
Create a strategy

Complete the square on both the x terms and the y terms.

Apply the idea
\displaystyle x^{2} + 4 x + y^{2} + 6 y - 3 \displaystyle =\displaystyle 0
\displaystyle x^{2} + 4 x + 2^2 + y^{2} + 6 y + 3^2- 3 \displaystyle =\displaystyle 2^2 + 3^2Add 2^2 and 3^2 to both sides
\displaystyle (x+2)^2 + (y+ 3)^2- 3 \displaystyle =\displaystyle 2^2 + 3^2Write in factorised form
\displaystyle (x+2)^2 + (y+ 3)^2- 3 + 3 \displaystyle =\displaystyle 2^2 + 3^2 +3Add 3 to both sides
\displaystyle (x+2)^2 + (y+ 3)^2\displaystyle =\displaystyle 16Evaluate
b

What are the coordinates of the centre of this circle?

Worked Solution
Create a strategy

Equations of the form (x-h)^{2} + (y-k)^{2} = r^{2} are circles with centre at (h,k).

Apply the idea

In part (a) we have the equation \left(x+2\right)^{2} + \left(y+ 3\right)^{2} = 16 which can be written as \\ \left(x - (-2)\right)^{2} + \left(y - (-3)\right)^{2} = 16 and it is now of the form (x-h)^{2} + (y-k)^{2} = r^{2}. So we have h=-2 and k=-3.

The centre of the circle is (-2, -3).

c

What is the radius of this circle?

Worked Solution
Create a strategy

Equations of the form (x-h)^{2} + (y-k)^{2} = r^{2} are circles with radius of r.

Apply the idea

In part (b) the equation (x+2)^{2} + (y+3)^{2} = 16 can be written in the form (x-h)^{2} + (y-k)^{2} = r^{2}, we can say that:

\displaystyle r^{2}\displaystyle =\displaystyle 16
\displaystyle \sqrt{r^{2}}\displaystyle =\displaystyle \sqrt{16}Square root both sides
\displaystyle r\displaystyle =\displaystyle 4Evaluate

The radius of the circle is 4 units.

Idea summary

The graph of an equation of the form \left(x-h \right)^{2} + \left(y-k \right)^{2} = r^{2} is a circle.

Circle have a centre at (h,k) and a radius of r.

Circles of the form x^{2} + y^{2} = r^2 can be vertically translated by k units up get: \\ x^{2} + \left(y-k \right)^{2} = r^2 or k units down to get: x^{2} + \left(y-k \right)^{2} = r^2.

Circles of the form x^{2} + y^{2} = r^2 can be horizontally translated by h units to the left to get: \left(x+h \right)^{2}+ y^{2} = r^2 or h units to the right to get: \left(x-h \right)^{2}+ y^{2} = r^2.

Circles of the form x^{2} + y^{2} = 1 can be scaled by a scale factor of r to get the equation x^{2} + y^{2} = r^{2}.

The general equation of a circle is given by:

\displaystyle (x-h)^2+(y-k)^2=r^2
\bm{(h,k)}
are the coordinates of the centre of the circle
\bm{r}
is the radius of the circle

Outcomes

MA5.1-7NA

graphs simple non-linear relationships

MA5.2-10NA

connects algebraic and graphical representations of simple non-linear relationships

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