CIRCLES AND SEMI-CIRCLES
Example 8:
Given the equation of a circle
its radius. Also find the coordinates of the intercepts.
, find the coordinates of its center and
Finding the coordinates of the center and the radius:
To find this information we must change the equation to standard form
.
To achieve this, we must complete the square on both x and y. There is a review of square
completion below the Lecture 03 notes!
NOTE that the coefficient of
and
is 3 and not 1 as shown in previous
examples. Please understand that as long as the coefficients are the same number
we still get the graph of a circle. Only if the coefficients are different would we get a
different picture.
First, we will group all terms. Note that we can write
as
!!!
We'll need to find the term missing in the parentheses containing y-values to allow us to find a
perfect square trinomial. We will use the coefficient of the y-term, which is - 4, divide it by 2,
and raise this result to the second power.
This results in
.
Now, let's add 4 into the parentheses containing y-values on the left side of the equal sign.
Please note, that we actually added 3(4) = 12 due to the 3 in front of the y-values. It actually
triples the values in the parentheses. Finally, to preserve equality we therefore must also add
12 to the right side of the equal sign.
and factoring and combining like terms produces
.
You must now convert the above equation to standard form
the coefficients become 1 by dividing both sides by 3.
. That is,
We can now see that the coordinates of the center of the circle are (0, 2) and the radius
is 2.
Finding the coordinates of the intercepts:
Let's use the equation of the circle in standard form. Using it to find the coordinates of the intercepts
is much easier than trying to do the same given the general form.
We have
.
Coordinates of the x-intercept:
We know that x-intercepts are found by setting y = 0.
Now we use the Square Root Property to find that
equal to 0.
. Since -0 does not exist, x is simply
Therefore, the coordinates of the x-intercept (there is only one) is (0, 0).
Coordinates of the y-intercept:
We know that y-intercepts are found by setting x = 0.
Now we use the Square Root Property to find that
and
Therefore, y - 2 = 2 and y - 2 = -2, so that y = 4 and y = 0.
Therefore, the coordinates of the y-intercepts are (0, 4) and (0, 0).
.