UK Secondary (7-11)

Circles (mixed set)

Lesson

In The Origin of Circles, we looked at the general equation of a circle whose centre point is at the origin. This equation was: $x^2+y^2=r^2$`x`2+`y`2=`r`2.

The standard form of a circle with centre coordinates $\left(h,k\right)$(`h`,`k`) and radius, $r$`r`, is:

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

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.

Remember!

Here are the standard and general forms of the equation of a circle.

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

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

Centre at Origin then: $x^2+y^2=r^2$`x`2+`y`2=`r`2

As we already know, a semicircle is half of a circle.

The semicircle equations are really useful as they split the circle relation, into two semicircular functions.

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(`x`−`h`)2+(`y`−`k`)2=`r`2 is the equation of a circle. To turn this into two semicircular functions we solve for y.

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

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

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

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

Let's just look at what happens if we plot the two parts of this solution separately.

Use this interactive and select the box to switch between the positive solution and the negative solution. Can you see what happens?

What you have just discovered is that the positive part $y=\sqrt{r^2-\left(x-h\right)^2}+k$`y`=√`r`2−(`x`−`h`)2+`k` yields the top half of a semicirlce, and the negative part, $y=-\sqrt{r^2-\left(x-h\right)^2}+k$`y`=−√`r`2−(`x`−`h`)2+`k` yields the bottom half of the semicirlce.

State the equation of the circle.

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

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

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

c) Graph the circle.