Circles and Ellipses

Lesson

The diagram below shows that the domain of a circle consists of all $x$`x`-values within the interval $a-r\le x\le a+r$`a`−`r`≤`x`≤`a`+`r` and the range of of a circle consists of all $y$`y`-values within the interval $b-r\le y\le b+r$`b`−`r`≤`y`≤`b`+`r`.

We can also write the domain in interval notation as $\left[a-r,a+r\right]$[`a`−`r`,`a`+`r`], and the range in interval notation as $\left[b-r,b+r\right]$[`b`−`r`,`b`+`r`].

The circle itself is not a function, but can be split into two semicircles, each of which are functions. We can rearrange the equation to make this clear:

$\left(x-a\right)^2+\left(y-b\right)^2$(x−a)2+(y−b)2 |
$=$= | $r^2$r2 |

$\left(y-b\right)^2$(y−b)2 |
$=$= | $r^2-\left(x-a\right)^2$r2−(x−a)2 |

$y-b$y−b |
$=$= | $\pm\sqrt{r^2-\left(x-a\right)^2}$±√r2−(x−a)2 |

$y$y |
$=$= | $y=b\pm\sqrt{r^2-\left(x-a\right)^2}$y=b±√r2−(x−a)2 |

So, for example, in the new form the circle whose centre is located at $\left(2,5\right)$(2,5) and has radius $r=3$`r`=3 has the equation given by $y=5\pm\sqrt{9-\left(x-2\right)^2}$`y`=5±√9−(`x`−2)2. It can be split into the two functions, say $f$`f` and $g$`g`, where $f\left(x\right)=5+\sqrt{9-\left(x-2\right)^2}$`f`(`x`)=5+√9−(`x`−2)2 and $g\left(x\right)=5-\sqrt{9-\left(x-2\right)^2}$`g`(`x`)=5−√9−(`x`−2)2.

Each of these functions are semicircles. Both functions will have the same domain; that of the original circle, given by $-1\le x\le5$−1≤`x`≤5. The range of $f$`f` becomes $5\le y\le8$5≤`y`≤8 and the range of $g$`g` becomes $2\le y\le5$2≤`y`≤5 as shown here.

Consider the graph of the circle shown below.

Loading Graph...

State the domain of the graph in interval notation.

State the range of the graph in interval notation.

Consider the equation $\left(x+5\right)^2+\left(y+3\right)^2=16$(`x`+5)2+(`y`+3)2=16.

Plot the graph described by the equation.

Loading Graph...State the domain of the graph in interval notation.

State the range of the graph in interval notation.

The top of a semicircle has a domain of $\left[-10,2\right]$[−10,2] and a range of $\left[-2,4\right]$[−2,4].

Plot the semicircle.

Loading Graph...State the equation for the semicircle in the form $y=\pm\sqrt{r^2-\left(x-h\right)^2}+k$

`y`=±√`r`2−(`x`−`h`)2+`k`.

Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems