Hong Kong

Stage 4 - Stage 5

Lesson

The following interactive allows you to explore the standard form equation of a circle. It shows how the equation changes as the radius $r$`r` changes. To change the radius, just drag the $r$`r` slider.

General Equation for Circle with Centre at the Origin

$x^2+y^2=r^2$`x`2+`y`2=`r`2

where $\left(x,y\right)$(`x`,`y`) are a pair of coordinates and $r$`r` is the radius

*Proof:*

This formula is derived using Pythagoras' theorem. Consider the following graph, which is a circle with the centre at $\left(0,0\right)$(0,0) and a radius of $5$5 units. The blue radius touches the circle at $\left(4,3\right)$(4,3).

Let's draw in a right-angled triangle:

Using Pythagoras' theorem:

$3^2+4^2$32+42 | $=$= | $9+16$9+16 |

$=$= | $25$25 | |

$=$= | $r^2$r2 |

$\therefore$∴ $x^2+y^2=r^2$`x`2+`y`2=`r`2

State the equation of the circle.

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Consider the circle $x^2+y^2=4$`x`2+`y`2=4.

Find the $x$

`x`-intercepts. Write all solutions on the same line separated by a comma.Find the $y$

`y`-intercepts. Write all solutions on the same line separated by a comma.Graph the circle.

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Circles that are centred at the origin have the equation $x^2+y^2=r^2$`x`2+`y`2=`r`2.

Below is a circle with radius $3$3 units that is centred at the origin.

How would the equation change with vertical translations of this circle?

By vertically translating a circle, we are moving it above or below the origin. In the applet below, click and hold the centre of the circle to drag it up and down along the $y$`y`-axis. You can also move the circle vertically with the $k$`k` slider, and adjust the radius with the $r$`r` slider.

Think about the following questions:

- When we change the circle's radius, which part of the equation is affected?
- When we translate the circle vertically, which part of the equation is affected?
- What is the relationship between the $y$
`y`-coordinate of the circle's centre $k$`k`and the equation?

By using the applet above, we can make the following observations:

- The term on the right side of the equation is always the square of the radius $r$
`r`. - By translating the circle vertically $k$
`k`units from the origin, its centre has the coordinates $\left(0,k\right)$(0,`k`). - $k$
`k`is positive when the circle's centre is above the $x$`x`-axis and negative when it is below. - The equation takes the form $x^2+\left(y-k\right)^2=r^2$
`x`2+(`y`−`k`)2=`r`2 with a vertical translation from the origin.

Vertical translations of circles

The equation of a circle translated $k$`k` units vertically from the origin is

$x^2+\left(y-k\right)^2=r^2$`x`2+(`y`−`k`)2=`r`2,

where $r$`r` is the radius of the circle.

Consider the graph of the circle shown in the Diagram 1.

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Diagram 1 |

The graph of the circle is translated to the position shown in Diagram 2.

Loading Graph...Diagram 2 Which of the following statements is true?

The graph of the circle has been translated $6$6 units up.

AThe graph of the circle has been translated $6$6 units down.

B

Which of the following equations describe a circle that has been translated $2$2 units downwards from the origin?

$x^2+\left(y+2\right)^2=r^2$

`x`2+(`y`+2)2=`r`2A$\left(x-2\right)^2+y^2=r^2$(

`x`−2)2+`y`2=`r`2B$\left(x+2\right)^2+y^2=r^2$(

`x`+2)2+`y`2=`r`2C$x^2+\left(y-2\right)^2=r^2$

`x`2+(`y`−2)2=`r`2D

Given the graph of $x^2+y^2=6^2$`x`2+`y`2=62, draw the graph of $x^2+\left(y+3\right)^2=6^2$`x`2+(`y`+3)2=62.

- Loading Graph...

Circles that are centred at the origin have the equation $x^2+y^2=r^2$`x`2+`y`2=`r`2.

Below is a circle with radius $3$3 units that is centred at the origin.

How would the equation change with horizontal translations of this circle?

By horizontally translating a circle, we are moving it to the left or right of the origin. In the applet below, click and hold the centre of the circle to drag it left and right along the $x$`x`-axis. You can also move the circle horizontally with the $h$`h` slider, and adjust the radius with the $r$`r` slider.

Think about the following questions:

- When we change the circle's radius, which part of the equation is affected?
- When we translate the circle horizontally, which part of the equation is affected?
- What is the relationship between the $x$
`x`-coordinate of the circle's centre $h$`h`and the equation?

By using the applet above, we can make the following observations:

- The term on the right side of the equation is always the square of the radius $r$
`r`. - By translating the circle horizontally $h$
`h`units from the origin, its centre has the coordinates $\left(h,0\right)$(`h`,0). - $h$
`h`is positive when the circle's centre is on the right hand side of the $y$`y`-axis and negative when it is on the left. - The equation takes the form $\left(x-h\right)^2+y^2=r^2$(
`x`−`h`)2+`y`2=`r`2 with a horizontal translation from the origin.

Horizontal translations of circles

The equation of a circle translated $h$`h` units horizontally from the origin is

$\left(x-h\right)^2+y^2=r^2$(`x`−`h`)2+`y`2=`r`2,

where $r$`r` is the radius of the circle.

Consider the following circles with their respective equations below them.

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Loading Graph... |

x2+y2=9 |
x−2)2+y2=9 |

The graph of $\left(x-2\right)^2+y^2=9$(

`x`−2)2+`y`2=9 has been horizontally translated from the graph of $x^2+y^2=9$`x`2+`y`2=9 by $2$2 units to the:Left

ARight

B

Consider the following circles with their respective equations below them.

Loading Graph... |
Loading Graph... |

x2+y2=4 |
x+3)2+y2=4 |

The graph of $\left(x+3\right)^2+y^2=4$(

`x`+3)2+`y`2=4 has been horizontally translated from the graph of $x^2+y^2=4$`x`2+`y`2=4 by $3$3 units to the:Left

ARight

B

Consider a circle of radius $3$3 units, centred at the origin. If this circle was horizontally translated $4$4 units to the right of the origin, what would be its equation?

$x^2+\left(y-4\right)^2=9$

`x`2+(`y`−4)2=9A$\left(x+4\right)^2+y^2=9$(

`x`+4)2+`y`2=9B$x^2+\left(y+4\right)^2=9$

`x`2+(`y`+4)2=9C$\left(x-4\right)^2+y^2=9$(

`x`−4)2+`y`2=9D

The following interactive allows you to explore the standard form equation of a circle. It shows how the equation changes as the coordinates of the centre ($h$`h`, $k$`k`) and the radius $r$`r` change. To move the circle, drag the sliders for $h$`h` and $k$`k`, the centre coordinates of the circle, while to change the radius, just drag the $r$`r` slider. You'll notice that, regardless of the values, the equation will always be in the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(`x`−`h`)2+(`y`−`k`)2=`r`2.

Equation of a Circle in Standard Form

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(`x`−`h`)2+(`y`−`k`)2=`r`2

where $\left(h,k\right)$(`h`,`k`) is the coordinates of the centre of the circle

and $r$`r` is the radius of the circle

Equation of a Circle in General Form

The standard form of a circle can also be rearranged and written in general form. The general form of a circle is:

$x^2+y^2+ax+by+c=0$`x`2+`y`2+`a``x`+`b``y`+`c`=0

A circle is described by the following equation:

$\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{9}{4}$(`x`+12)2+(`y`+12)2=94

Find the centre of the circle.

Find the radius of the circle.

Hence plot the graph for the given circle.

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Consider the circle whose equation is $\left(x+1\right)^2+\left(y+4\right)^2=25$(`x`+1)2+(`y`+4)2=25.

State the radius.

State the coordinates of the centre of the circle.

Hence, graph the circle on the coordinate plane below.

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We can rewrite the equation of the circle by making $y$`y` the subject as follows.

$\left(x-h\right)^2+\left(y-k\right)^2$(x−h)2+(y−k)2 |
$=$= | $r^2$r2 |
(Writing the equation) |

$\left(y-k\right)^2$(y−k)2 |
$=$= | $r^2-\left(x-h\right)^2$r2−(x−h)2 |
(Subtracting by $\left(x-h\right)^2$(x−h)2) |

$y-k$y−k |
$=$= | $\pm\sqrt{r^2-\left(x-h\right)^2}$±√r2−(x−h)2 |
(Taking the square root) |

$y$y |
$=$= | $\pm\sqrt{r^2-\left(x-h\right)^2}+k$±√r2−(x−h)2+k |
(Adding $k$k) |

We can see that the resulting equation contains two parts:

$y$y |
$=$= | $\sqrt{r^2-\left(x-h\right)^2}+k$√r2−(x−h)2+k |
(1) |

$y$y |
$=$= | $-\sqrt{r^2-\left(x-h\right)^2}+k$−√r2−(x−h)2+k |
(2) |

We might be curious to find out which sections of the graph represent each part of the equation.

If $x=h$`x`=`h`, then:

$y$y |
$=$= | $\sqrt{r^2-\left(x-h\right)^2}+k$√r2−(x−h |