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.

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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.

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).

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.

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,

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 16	Simplify
\displaystyle x^2 -6x + 9 + y^2 + 4y + 4	\displaystyle =	\displaystyle 16	Expand the perfect squares
\displaystyle x^2 -6x + y^2 + 4y + 13	\displaystyle =	\displaystyle 16	Combine like terms
\displaystyle x^2 -6x + y^2 + 4y + 13 -16	\displaystyle =	\displaystyle 16 -16	Subtract 16 from both sides
\displaystyle x^2 -6x + y^2 + 4y -3	\displaystyle =	\displaystyle 0	Evaluate
\displaystyle x^2 + y^2 -6x +4y -3	\displaystyle =	\displaystyle 0	Write in general form

Example 3

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

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 + 3^2	Add 2^2 and 3^2 to both sides
\displaystyle (x+2)^2 + (y+ 3)^2- 3	\displaystyle =	\displaystyle 2^2 + 3^2	Write in factorised form
\displaystyle (x+2)^2 + (y+ 3)^2- 3 + 3	\displaystyle =	\displaystyle 2^2 + 3^2 +3	Add 3 to both sides
\displaystyle (x+2)^2 + (y+ 3)^2	\displaystyle =	\displaystyle 16	Evaluate

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).

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 4	Evaluate

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

VCMNA339

Explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital technology as appropriate

VCMNA359 (10a)

Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations.

5.09 Circles

Ideas

Transformations of circles

Exploration

Examples

Example 1

Example 2

Example 3

Outcomes

VCMNA339

VCMNA359 (10a)

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