Lesson

When we looks at graphs of functions or relations, we concentrate our focus on key characteristics of a graph.

We're interested in examining the graph of a circle, and to do that we need to concentrate our focus on two key characteristics.

When describing a circle on the $xy$`x``y`-plane, we want to highlight two key features. As when we were finding the area and circumference of circles, the radius is a key characteristic. The center of the circle is the other defining feature of a circle graph.

On the graph below, you can see, by inspection, that the center of the circle is $(2,-5)$(2,−5)

We can determine the radius of the circle by counting the number of units from the center, out to the left or right of the center, or directly up or down from the center. So for this graph the radius is $4$4 units.

Key features of a circle graph

center: The point which is the same distance from every point on the circle, written as an ordered pair $\left(h,k\right)$(`h`,`k`)

radius: The distance from the center to any point on the circle

Consider the graph of the circle shown below.

Loading Graph...

Complete the statement.

Every point on the circle is exactly $\editable{}$ units away from the point

$($($\editable{}$, $\editable{}$$)$).

Consider the circle in the graph.

Loading Graph...

State the coordinates of the center in the form $\left(a,b\right)$(

`a`,`b`).State the diameter.

We want to determine the formula for the equation of a circle with its center at the origin, $\left(0,0\right)$(0,0), then we will look at transformations.

Recall the Pythagorean theorem which states that $a^2+b^2=c^2$`a`2+`b`2=`c`2 where $c$`c` is the length of the hypotenuse of a triangle and $a$`a` and $b$`b` are the lengths of the other two sides. Consider the following graph, which is a circle with the center 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 triangle:

Using Pythagoras' theorem:

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

$=$= | $25$25 | |

$=$= | $5^2$52 |

So we have $3^2+4^2=5^2$32+42=52, but $\left(3,4\right)$(3,4) was a point $\left(x,y\right)$(`x`,`y`) on the circle and $5$5 is the radius, $r$`r`, of the circle, so we get:

$3^2+4^2$32+42 | $=$= | $5^2$52 |

$x^2+y^2$x2+y2 |
$=$= | $r^2$r2 |

Standard equation for circle with center at the Origin

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

where $\left(x,y\right)$(`x`,`y`) is any point on the circle and $r$`r` is the radius

State the equation of the circle.

Loading Graph...

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.

Loading Graph...

Now that we know $x^2+y^2=r^2$`x`2+`y`2=`r`2 represents a circle with center at the origin, $\left(0,0\right)$(0,0), and radius, $r$`r`, we can explore circles not centered at the origin. 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 center $\left(h,k\right)$(`h`,`k`) and the radius $r$`r` change. To move the circle, drag the sliders for $h$`h` and $k$`k`, the center coordinates of the circle, while to change the radius, just drag the $r$`r` slider.

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 center of the circle

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

The equation of a circle is given by $\left(x+4\right)^2+\left(y-2\right)^2=25$(`x`+4)2+(`y`−2)2=25.

Find the coordinates of the center of this circle.

What is the radius of the circle?

Plot the graph for the given circle.

Loading Graph...

A circle has its center at ($3$3,$3$3) and a radius of 6 units.

Plot the graph for the given circle.

Loading Graph...Write the equation of the circle.

Recall that a circle is the set of all points equidistant from the same point - the center of the circle. So every point on a circle is the same distance from its center.

Observe what happens as you change the following in the applet below:

- Keep the slider the same value as the radius (set $d=4$
`d`=4). Explain the location of point $B$`B`as you move it around. - Now move the slider to a value greater than the radius. Explain the location of point $B$
`B`relative to the circle. - Now move the slider to a value less than the radius. Explain the location of point $B$
`B`relative to the circle.

Hopefully, you were able to notice the following

When $d=r$`d`=`r`, point $B$`B` lies on the circle

When $d>r$`d`>`r`, point $B$`B` lies outside the circle

When $d`d`<`r`, point $B$`B` lies inside the circle

How do you determine the distance from a point $B$`B`$\left(x,y\right)$(`x`,`y`) to the center of the circle $A$`A`$\left(h,k\right)$(`h`,`k`)?

The same way we would determine the distance between any two points. Using the distance formula or the Pythagorean theorem.

Since the radius of a circle is always positive, a little substitution gives the following:

Summary

Where $\left(x,y\right)$(`x`,`y`) is a point, $B$`B`, on the $xy$`x``y`-plane,

- If $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(
`x`−`h`)2+(`y`−`k`)2=`r`2, point B lies on the circle. - If $\left(x-h\right)^2+\left(y-k\right)^2>r^2$(
`x`−`h`)2+(`y`−`k`)2>`r`2, point B lies outside the circle. - If $\left(x-h\right)^2+\left(y-k\right)^2
( `x`−`h`)2+(`y`−`k`)2<`r`2, point B lies inside the circle.

We can use these formulas to prove a point's location algebraically.

Determine whether the following points lie inside, outside, or on the circle $x^2+y^2=117$`x`2+`y`2=117.

$\left(-8,5\right)$(−8,5)

Inside

ACan't say

BOn

COutside

DInside

ACan't say

BOn

COutside

D$\left(-11,11\right)$(−11,11)

Outside

AInside

BOn

CCan't say

DOutside

AInside

BOn

CCan't say

D$\left(0,0\right)$(0,0)

Inside

AOutside

BCan't say

COn

DInside

AOutside

BCan't say

COn

D

Derive the equation of a circle of given center and radius using the Pythagorean Theorem. Find the center and radius of a circle, given the equation of the circle.