12.08 Equations of circles

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.

Characteristics of a circle graph

When describing a circle on the $xy$xy-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.

 

Practice questions

QUESTION 1

QUESTION 2

 

Equation of a circle with center $\left(0,0\right)$(0,0)

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.

Exploration

Recall the Pythagorean theorem which states that $a^2+b^2=c^2$a2+b2=c2 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

 

Practice questions

Question 3

Question 4

 

Equation of a circle with any center

Now that we know $x^2+y^2=r^2$x2+y2=r2 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. 

 

 

Practice questions

Question 5

Question 6

 

Inside, outside or on a 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.  

Exploration

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

1. 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.
2. Now move the slider to a value greater than the radius.  Explain the location of point $B$B relative to the circle.
3. 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 $dd<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:

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

Practice question

Question 7