Example
1 Write down the equation of a circle with radius 8 and center (−4,7) .
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Solution
Okay,
in this case we have r = 8, h
= −4 and k = 7 so all we
need to do is plug them into the standard
form of the equation of the circle.
( x – (–4))2 + (y
– 7)2 = 82
( x + 4)2 + (y – 7)2 = 82
Example 2 Determine the center and
radius of each of the following circles and sketch the graph of the circle.
(a) x2 + y2 =1
(b) x2 + ( y −3)2 = 4
(c) (x −1)2 + (y + 4)2 = 16
Solution
In all of these all that we really need to do is compare the
equation to the standard form and identify the radius and center. Once that is done find the four
points talked about above and sketch in the circle.
(a) x2 + y2 = 1
In this case it’s just x and y squared by
themselves. The only way that we could have this is to have both h and k be zero. So, the center and radius
is,
center = (0,0) radius = √1 =1
Don’t forget that the radius is the square root of the number on
the other side of the equal sign.
Here is a sketch of this circle.
A circle centered at the origin
with radius 1 (i.e. this circle) is called the unit circle. The
unit circle is very useful in a Trigonometry class.
(b) x2 + ( y −3)2 = 4
In this part, it looks like the x coordinate of the center
is zero as with the previous part. However, this time there is something more
with the y term and so comparing this term to the standard form of the circle we can see that the y coordinate of the center must be 3. The
center and radius of this circle is then,
center = (0,3) radius = √4 = 2
Here is a sketch of the circle. The center is marked with a red
cross in this graph.
(c) (x −1)2 + (y + 4)2 = 16
For
this part neither of the coordinates of the center are zero. By comparing our
equation with the standard form it’s fairly easy to see (hopefully…) that the x
coordinate of the center is 1. The y coordinate isn’t too bad
either, but we do need to be a little careful. In this case the term is (y + 4)2 and
in the standard form the term is (y − k)2.
Note that the signs are different. The only way that this can happen is if k
is negative. So, the y coordinate of the center must be -4.
The
center and radius for this circle are,
center = (1,4) radius
= √16 = 4
Here
is a sketch of this circle with the center marked with a red cross.
Example 3 Determine the center and
radius of each of the following.
(a) x2 + y2 +8x
+ 7 = 0
(b) x2 + y2 − 3x
+10y − 1 = 0
Solution
Neither of these equations are in standard form and so to
determine the center and radius we’ll need to put it into standard form. We actually already know how to
do this. Back when we were solving quadratic equations we saw a way to turn a
quadratic polynomial into a perfect square. The process was called completing the square.
This is exactly what we want to do here, although in this case we
aren’t solving anything and we’re going to have to deal with the fact that
we’ve got both x and y in the equation. Let’s step through the process with the first part.
(a) x2 + y2 +8x + 7 = 0
We’ll go through the process in a step by step fashion with this
one.
Step 1 : First get the constant
on one side by itself and at the same time group the x terms together
and the y terms together.
x2 +8x + y2 = −7
In this case there was only one term with a y in it and two
with x’s in them.
Step
2 : For each variable with two terms complete the square on those
terms.
So, in this case that means that we only need to complete the
square on the x terms. Recall how this is done. We first take half the coefficient of the x and
square it.
We then add this to both sides of the equation.
x2 +8x +16 + y2 = −7 +16 = 9
Now, the first three terms will factor as a perfect square.
(x + 4)2 + y2 = 9
Step 3 : This is now the standard
form of the equation of a circle and so we can pick the center
and radius right off this. They are,
center = (−4,0) radius = √9 = 3
(b) x2 + y2 − 3x
+10y − 1 = 0
In this part we’ll go through the process a little quicker. First
get terms properly grouped and
placed.
Now, as noted above we’ll need to complete the square twice here,
once for the x terms and once for the y terms. Let’s first get
the numbers that we’ll need to add to both sides.
Now, add these to both sides of the equation.
When adding the numbers to both sides make sure and place them
properly. This means that we need to put the number from the coefficient of the
x with the x terms and the number from the coefficient of the y
with the y terms. This placement is important since this will be the
only way that the quadratics will factor as we need them to factor.
Now, factor the quadratics as show above. This will give the standard
form of the equation of the circle.
This looks a little messier than the equations that we’ve seen to
this point. However, this is
something that will happen on occasion so don’t get excited about
it. Here is the center and
radius for this circle.
Do
not get excited about the messy radius or fractions in the center coordinates.
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