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Hello, and welcome to Bay College's online lecture
series for College Algebra.
This a section 2.4.
And it deals with circles.
The first thing we're going to consider is the point h, k
where this is just some x value and some y value at this
point and some arbitrary other point, x,y.
Now they live r units away from one another.
So let's use the distance formula as we had seen in
previous videos.
If we have the distance formula.
I'm going to take the square root of my changes in x plus
my change in y.
And these units are squared.
This is going to equal, in this case, r because the
distance between these points.
And we're calling it r.
So what is my change in x?
Well, I have this x value minus this x value.
This y value here minus this y value.
Now to eliminate this square root, I'm just going to square
both sides.
And if I square both sides, the square root goes away.
This is the equation of a circle.
A circle is defined as all the points in a plane that are r
units away from the center.
So h,k actually describes the center of a circle.
And if I go r units away here or r units away here or here
or here or all the way around, we actually see that we're
actually making a circle from this radius.
And I'll just continue it all the way around.
Hopefully, my free hand circle doesn't look too bad.
But what this describes is a circle, any point on here.
Now, hopefully, we realize that the equation of a circle
is actually very similar to the distance formula.
And as we had seen in previous videos, the distance formula
is a variation of Pythagorean theorem.
And I hope this doesn't blow your mind too much.
But circles and triangles, at least right triangles, are
actually described using the same equation.
Because essentially what we're looking at is any change in x
and change in y that describes a triangle.
What if my arbitrary x,y point was up here?
Well, I go x units this way, y units that way, and
this is always r.
It's a different triangle.
So if we have an infinite number of triangles, it's
described using this equation which looks very similar to
Pythagorean theorem.
Now one other thing we want to define is this right here.
If we have a circle that's centered at the origin, where
my h is 0 and k is 0 minus 0 minus 0 in those values.
We get what's called the unit circle if the radius is 1.
And the unit circle looks just like that, the unit circle is
centered at the origin.
h and k in this case are 0.
And r is 1.
Now when it comes to determining what h and k is,
always keep in mind that it's minus the h value and minus
the k value.
When they're in this grouping symbol before it's squared,
you're always going to see the opposite value you're looking
for within that symbol.
Let's actually look at an example of a circle.
If we come over here, we're asked to graph this circle.
And this circle is in standard form, just as we've see over
here. x minus h quantity squared plus y minus k
quantity squared equals the radius squared.
So our center is the h,k value.
Well, let's determine what h and k are.
Well, what am I subtracting from x before I square it?
Well, there's nothing being subtracted.
So h, in this case, is 0.
What am I subtracting from y before I square it?
Well, here as I said before, notice it's the opposite of
what I see in there.
Because it's y minus the k value.
Well, y minus a negative 3 would give me y plus 3.
So k, in this case, would be a negative 3.
So that's how we determine the circle when it's written in
standard form.
We just say what am I subtracting from x?
What am I subtracting from y?
It's always going to be the opposite of what I actually
see in there.
Now, the radius, we look at this value right here, what
it's set equal to.
Well, the radius, this is the radius squared.
So the radius, I'd have to take the square root of that.
Now when we introduced square roots, we think plus or minus.
But because the radius is defined as a distance, we only
have to worry about its absolute value.
It's going to be 2.
So the square root of 4 is 2.
Well, let's find some additional information.
We know how to find x-intercepts.
We just set the y value to 0.
If this is 0, 3 squared is 9.
Subtract 9 from both sides, and I get, I'm going to write
it right here, x squared equals negative 5.
And that's just setting y to 0 and getting my constants to
the other side.
Now to solve this, I take the square root of both sides.
Well, I realize right now if I take the square root of both
sides, I'm taking the square root of a negative number.
That's not a real value.
That's an imaginary value.
That tells me something about the x-intercepts.
That means there are no x-intercepts.
So when it comes to graphing it, I'm going to realize that
it will not cross the x-intercept or the x-axis
because there are no x-intercepts.
Let's find the y-intercept.
Well, we take the equation and we set x equal to zero.
Well, this we have a perfect square equal 4.
We can use the square root method.
So if I do the square root method, I'll write it right
here, y plus 3 equals plus or minus the square
root of 4 is 2.
And now I can just subtract 3 from both sides, y equals
negative 3 plus or minus 2.
Well, that's actually two separate values, negative 3
plus 2 and negative 3 minus 2 which tells me I have 2
y-intercepts.
In this case, negative 3 minus 2.
When x was 0, I get negative 5.
And when x is 0, negative 3 plus 2 negative 1.
Two y-intercepts, no x-intercepts.
Now with all this information, the center, the radius, any
intercepts that exist, I can go to my graph and graph it.
Now I said this is important because when I graph it, it
better not cross the x-axis.
So let's first start with our center.
Our center is 0, negative 3.
So x is 0 and our y value is negative 3.
And I know from this center I can go two in any direction.
Maybe I go two to the right, one, two.
Two to the left, one, two.
Two up and two down.
And if we notice, what were my y-intercepts?
0, negative 5.
Well, that's two up from this value-- or excuse me.
0, negative 1, that's two up from this value.
And the 0, negative 5 is two down from this value.
And now with enough points there.
I see I have four points.
Well, that should be enough to graph the circle.
And even though it's free handed, it
looks relatively circleish.
It should be a circle.
All right, what happens if our circles are
not in standard form?
Sometimes we'll see circles written in general form.
Well, let's just take for a moment and expand
what we just had here.
x squared plus y plus 3 squared equals 4.
What we're going to do is we're going to expand this to
put it into general form.
So we can see that they are very similar.
Well, they're actually the same just a different way of
looking at it.
Well, x squared is what it is.
And this I can use FOIL to FOIL it out.
I'm going to get y squared plus 6 y plus 9 equals to 4.
And in general form, notice it's set equal to 0.
Well, let me just subtract 4 from both sides.
And now it's in general form.
So we could say, yes, we have an x squared term, we have a y
squared term, a,x, well, we don't have an a,x term.
Well, that's because our a coefficient is actually 0.
0 times x, there's not going to be an x there.
But we do have our b,y.
And we do have a constant of 5.
Now notice this is not the same as that radius.
All right, so let's graph an equation in general form.
Well, in order to graph this, I want to have
it in standard form.
I want to have it in the quantity x minus h quantity
squared plus y minus k quantity squared equals to the
radius squared.
So in order to do that, we're going to have to do something
called completing the square.
We've reviewed completing the square in a previous video.
So, hopefully, we remember how to complete the square.
In order to do it on a circle, essentially what we have to do
is we don't just do it once, we do it twice.
And we start out by grouping our variables together.
x squared and then we have a minus 6 x.
So I group those together, and I'm going to leave a space
here because if we recall, when we complete the square,
we're going to add a coefficient to make this a
perfect square.
And then I'm going to have my y squared and my 2 y.
And I'm going to leave a space.
And then if this had a constant, I would just move it
across the equal sign.
But in this case, there is no constant like this
that I can move over.
I'm essentially moving 0.
Now we can complete the square.
Hopefully, we remember how to do that.
We take half of this quantity and square it.
Well, half of negative 6 is negative 3.
Negative 3 squared is a positive 9.
What you do to one side of an equation you must
always do to the other.
Don't forget that.
That's a common mistake that students do when using
completing the square.
Now we completed the square here, but we also
have to do it here.
Well, half of 2 is 1 and 1 squared is 1.
I added 1 to this side of the equation, I have to remember
to add 1 to that side of the equation.
Now we're ready to do a little bit of factoring and we'll
have the equation in standard form.
Well, this is a perfect square.
It factors to x and half of b.
In this case, x minus 3.
Plus this is now a perfect square of y and half of b
which was a positive 1 equals to 0 plus 9
plus 1 which is 10.
If we see it is now in standard form.
I can now determine what the center is, what the radius is,
and I can find any x-intercepts or y-intercepts.
So let's go ahead and do that because this did ask us to
graph this equation.
So we put it in standard form, we're ready to graph it.
So I'm just going to move over here.
Looking at this equation, my h,k is always the opposite of
what I see in here.
So instead of negative 3 it's positive 3.
Here I have y plus 1.
So I want negative 1, the opposite of
what I see in there.
And that is the center.
Then we'll look at the radius.
Well, the radius is always the square root of this value.
Well, the square root of 10, well, it's not a nice number.
It's an irrational number.
So let's just leave it as the square root of 10.
Let's find some x-intercepts.
Well, looking at this equation, hopefully, the
camera is following here.
The x-intercept, well, we know how to find an x-intercept.
We set y to 0.
So I have 1 squared is 1.
I can subtract 1 from both sides.
So I get x minus 3 quantity squared equals 9.
Well, that's nice.
It's a perfect square.
So I can use the square root method and get x equals 3 plus
or minus the square root of 9 which is three.
So 3 plus 3 is 6.
3 minus 3 is 0.
6 and 0 are my x-intercepts.
So 0, when y was 0 and 6 when y was 0, two different
intercepts.
Let's find any y-intercepts.
Well, that's just setting x to 0.
So if this is 0, negative 3 squared is 9.
Subtract 9 from both sides.
And I'll just write the equation over here. y plus 1
squared equals 1.
And what I can do here is I can use the square root method
and subtract 1.
Negative 1 plus or minus 1.
Well, the values I get here, negative 1 plus 1 is 0 and
negative 1 minus 1 is negative 2.
So I have two values.
I have when x is 0, I've got negative 2.
And when x was 0, I also got 0.
Well, this is actually good information.
This tells me something important.
x-intercept of 0,0 is the same as the y-intercept of 0,0.
Well, this is the origin.
That's where x and y actually share the same intercept.
All right, so let's take all this information and
put it on our graph.
All right we know the center is 3, negative 1.
So I'm going to go three in the x direction and
down one in the y.
And I'm going to label my center 3, negative 1.
Now the radius, I'm going to go square root of 10 units in
all directions from that.
Well, the square root of 10 is not a pretty number.
I know it's a little bit more than 3 because the square root
of 9 would be 3.
And this is more than 9.
Well, we'll off on that.
Look we have these three points to graph.
So that'll give us some idea.
And, hopefully, we'll see the pattern be able, yeah, I can
draw a circle with that given information.
Well, 0,0, that is an x and y-intercept.
I have 6,0.
So if I go over here 6 units, and then I have 0, negative 2.
So 0, negative 2.
And we can see that this going to be the square root of 10
units away.
And, hopefully, that's enough that we can say, all right
well, let's sketch the circle.
And when we're free handing circles, it's not always the
easiest thing to do.
But we get the idea.
It is kind of circleish.
Right.
Or maybe you'd want to use a protractor or something like
that, us super nerds have protractors.
We could then sketch that, and it would actually pass through
those points and give us all the rest as well.
All right, so here is your quiz.
I want you to try it for yourself.
Given this equation in general form, x squared plus y squared
minus 4x equals 0, write this equation in standard form.
Identify the center h and k, give the radius, any x or
y-intercepts, and then use that information
to graph the equation.
All right.
Let's move on to using graphing utilities.
At the college algebra level, you'll use graphing calculator
a lot more often than you would in any other class that
may have came before this.
So you have to know how to use your calculator.
Now some of you may know how to graph something like this,
this circle in a calculator by changing settings.
But to graph this in your most basic of graphing calculators,
in order to do that, you have to set this equal to y.
Now to solve this for y, essentially what I'd have to
do is subtract x squared from both sides.
So y squared equals negative x squared and this positive 9.
So I just move the x squared term across the equal sign.
Now to solve this for y, I've got to take the square root.
Now when I introduce a square root, I always remember it
could be plus or minus whatever I'm taking
the square root of.
Now our calculator, our standard calculators, aren't
going to accept plus or minus.
In order to graph this, when you go to your graphing
utility, you're going to have to put in the equation twice.
y equals the positive x squared plus 9.
Because we do have a domain restriction here, right.
And y equals the negative.
So in our calculator we have the opportunity to put in y1
or y2, the negative of our negative x squared plus 9.
Now if you put this into your standard graphing utility, you
will see this value when you graph it.
Because it is centered at the origin.
So we know that's our center.
You're going to have a radius of 3.
You're going to see something like this when this section of
the graph is put in.
On the other side, you're going to have to put in the
negative, and you'll see the bottom half of the circle.
So that's why you have plus or minus.
Now in some more advanced graphing utilities, you may
know how to set the parameters so that you can do
something like this.
OK.
Otherwise, you solve it for y, and you put it in each piece.
This is the top half of our circle.
This is the bottom half of our circle.
So why don't you take out your calculator and try that and
see that you can actually graph this.
All right.
So this has been Section 2.4 for College Algebra, Dealing
with Circles.
Thank you for watching.