### Video Transcript

Equation of a Circle

In this video, weβll discuss what a
circle is, how to derive the equation of a circle, and weβll learn how to find the
equation of a circle given its center and a point on the circle and how to find the
center and radius of a circle given its equation. Before we move on to finding the
equation of a circle, letβs briefly recall the mathematical definition of a
circle. A circle is the set or locus of all
points at given distance from a point. In other words, we have a center of
our circle, and every single point on our circle is the same distance away from the
center. We call this the radius of the
circle.

Now, to help us find an equation
for our circle, weβre going to need to plot it on a Cartesian graph. To do this, we need to pick a point
for our center; letβs choose the origin. This is the point with coordinates
zero, zero, and we represent this with an π. We want to find the equation of a
general circle, so weβll just refer to the radius as the letter π. So letβs pick a point on our
circle. Weβll call this the point π₯,
π¦. We need to find a relationship
between π₯ and π¦.

The first thing we can notice from
this sketch is that π¦ is not a function of π₯. The reason for this is consider an
input of a value of π₯. For π¦ to be a function of π₯, we
need one specific output for our function. However, we can see that there will
be two outputs in this case. Therefore, this canβt be
represented π¦ as a function of π₯, so we canβt use our normal method. Weβre going to try using a
geometric method. To start, weβll connect a vertical
line from our π₯-axis to the point π₯, π¦. And because this is a vertical
line, this means weβll get a right-angled triangle. In fact, we can find the height of
this triangle. Since this goes from the π₯-axis to
the point π₯, π¦, its height will be π¦.

Itβs also worth pointing out if our
point π₯, π¦ had been below the π₯-axis, we would instead need to take the absolute
value of π¦. So in both cases, this length will
just be the absolute value of π¦. We can, of course, do exactly the
same thing to find the width of our triangle. It goes from the π¦-axis to the
point π₯, π¦. So its width will be the absolute
value of π₯. We now have a right-angled triangle
where we know all three of the lengths. And weβre going to apply the
Pythagorean theorem to this triangle. Remember, this tells us the width
squared plus the height squared will be equal to the hypotenuse squared.

In our case, the width of our
triangle was equal to the absolute value of π₯. The height of our triangle was
equal to the absolute value of π¦. And the length of our hypotenuse
was equal to π. So by the Pythagorean theorem, we
must have the absolute value of π₯ squared plus the absolute value of π¦ squared is
equal π squared. But remember, it doesnβt matter if
a number is positive or negative when we square it; weβll still get the same
answer. So in actual fact, we can simplify
the absolute value of π₯ squared to be π₯ squared and the absolute value of π¦
squared to just be π¦ squared. So we can just simplify this
equation to get π₯ squared plus π¦ squared is equal to π squared. And this is the equation of our
circle centered at the origin of radius π.

Itβs also worth pointing out that,
technically, since our π₯- and π¦-intercepts do not form a right-angled triangle in
this way, we need to check these points satisfy our equation separately. However, using the radius of the
circle is equal to π, we can just point the coordinates and see that all of these
satisfy our equation. So every point on our circle
satisfies the equation π₯ squared plus π¦ squared is equal to π squared. Therefore, weβve derived that the
equation of a circle centered at the origin of a radius π is π₯ squared plus π¦
squared is equal to π squared.

But this then begs the question,
what would have happened if we didnβt choose the origin as our center? What if weβd picked a different
point? Well, we can do very much the same
thing. Letβs consider this example where
we pick the center to be the point β, π. Once again, weβll pick a point on
our circle and call it π₯, π¦. Remember, weβre having the radius
of our circle equal to π. And just like we did before, weβll
form the same right-angled triangle. This time we need to be a little
bit more careful when we find the length and width of our right-angled triangle. For example, to find the height of
our right-angled triangle, we go from the point with π¦-coordinate π to the point
with π¦-coordinate π¦. In other words, the height of this
right-angled triangle will be π¦ minus π.

But remember, sometimes our value
of π¦ will be smaller than our value of π. This would give us a negative
answer. So weβll take the absolute value of
π¦ minus π. And we can do something very
similar to find the width of our right-angled triangle. This time we go from β to π₯. This means the width of our
right-angled triangle will be the absolute value of π₯ minus β. And just like we did before, we now
apply the Pythagorean theorem to this right-angled triangle. This gives us the absolute value of
π₯ minus β squared plus the absolute value of π¦ minus π squared is equal to π
squared.

And remember, if weβre squaring an
absolute value sign, we donβt need the absolute value symbol. So weβll write this equivalently as
π₯ minus β all squared plus π¦ minus π all squared plus π squared. And this gives us an equation for
our circle centered at the point β, π with a radius of π. But remember, technically, there
are four points on our circle which donβt give us a right-angled triangle in this
form. But just as we did before, we can
use the fact that our radius is π to find the coordinates of each of these
points. We can then see that these also
satisfy our equation. So weβve shown every point on our
circle satisfies this equation.

Therefore, weβve shown a circle
centered at the point β, π with a radius π will have the equation π₯ minus β all
squared plus π¦ minus π all squared is equal to π squared. So given the center of our circle
and the radius of our circle, we can find the equation of the circle. But the opposite is also true. If weβre given the equation of our
circle, we can just find the center and the radius. Before we move on to some examples,
thereβs one more thing we need to talk about: the general form for the equation of a
circle. To do this, we need to distribute
the squares over our parentheses in our equation for a circle. Doing this, we get π₯ squared minus
two βπ₯ plus β squared plus π¦ squared minus two ππ¦ plus π squared is equal to π
squared.

But remember, β, π, and π are all
just constants, so we can call negative two β π, negative two π π, and β squared
plus π squared minus π squared π. So by using these and then
rearranging, we get π₯ squared plus π¦ squared plus ππ₯ plus ππ¦ plus π is equal
to zero. This is called the general form for
the equation of a circle. Letβs now move on to an example of
how to find the center and radius of a circle given its equation.

Find the center and radius of the
circle π₯ plus four squared plus π¦ minus two squared is equal to 225.

Weβre given the equation for a
circle. We need to use this to find the
center of our circle and the radius of our circle. To start, letβs recall the equation
of a circle. We know a circle with a center at
the point β, π and a radius of π will have the equation π₯ minus β all squared
plus one minus π all squared is equal to π squared. And we can see the equation weβre
given is almost in this form. We do have to be careful,
however. For example, weβre not subtracting
a constant from π₯; weβre adding the constant four. But remember, adding four is the
same as subtracting negative four. So we can in fact write this as π₯
minus negative four all squared plus π¦ minus two all squared is equal to 225.

Now, itβs really easy to see the
center of our circle. Our value of β is negative four,
and our value of π is two. All we have to do now is find the
radius of our circle. In this case, the radius squared
will be equal to 225. So we want π squared is equal to
225. Thereβs a few different ways of
doing this. For example, we could take the
square roots of both sides of this equation. Normally, we would get a positive
and a negative square root. But remember, in this case, this
represents the radius. This is a length, so it must be
positive. So we get that π is equal to the
positive square root of 225. We can calculate this; itβs equal
to 15. So we can write 225 as 15
squared. This means the radius of our circle
must be equal to 15.

Remember, the center of our circle
will be the point β, π. Weβve shown that β is equal to
negative four and π is equal to two. And of course, we already showed
the radius was 15. Therefore, given the equation of
the circle π₯ plus four all squared plus π¦ minus two all squared is equal to 225,
we were able to show the center of this circle was the point negative four, two and
the radius of this circle was 15.

Letβs now move on to an example
where weβre given the graph of a circle and we need to find the equation of this
circle.

In the figure below, find the
equation of the circle.

Weβre given the graph of a
circle. We need to find the equation of
this circle. Letβs start by recalling what we
know about the equation of a circle. We know that a circle centered at
the point β, π and that has a radius of π will have the equation π₯ minus β all
squared plus π¦ minus π all squared is equal to π squared. So to find the equation of a
circle, we just need to find its center and its radius. Remember, every point on our circle
will be equidistant from the center of our circle. In our case, the center of our
circle is labeled. We just need to find the
coordinates of this circle.

When moving up in a vertical
direction to our π₯-axis, we can see the π₯-coordinate of this center is negative
five. And by doing the same in the
horizontal direction, we can see the π¦-coordinate of this center is negative
four. So the center of our circle is
negative five, negative four. But how are we going to find the
radius of our circle? Remember, this will be the length
of any line from the center of our circle to our circle. Thereβs a lot of different choices
for which radius we can pick. One such example is to pick the
following line. We can see this is a horizontal
line which goes from the center of our circle to the point with π₯-coordinate
zero. In other words, our radius will be
the length of the horizontal line from π₯ is equal to negative five to π₯ is equal
to zero.

Of course, the length of this line
is just equal to five. So our value of π is five. We now need to substitute our
values of β, π, and π into our equation for a circle. Substituting in β is equal to
negative five, π is equal to negative four, and π is equal to five, we get π₯
minus negative five all squared plus π¦ minus negative four all squared is equal to
five squared. And we could leave our answer like
this. However, we can also simplify π₯
minus negative five to π₯ plus five and π¦ minus negative four to π¦ plus four. And of course we, can also evaluate
five squared to give us 25. Therefore, we were able to show the
equation of the circle given to us in the figure is π₯ plus five all squared plus π¦
plus four all squared is equal to 25.

Letβs now look at an example of
finding the equation of a circle in the general form.

Write, in the form ππ₯ squared
plus ππ¦ squared plus ππ₯ plus ππ¦ plus π is equal to zero, the equation of the
circle of radius 10 and center four, negative seven.

The question wants us to find the
equation of a circle of radius 10 and center four, negative seven. In particular, we can see weβre
asked to give this in the general form of a circle. To do this, weβll start with the
equation of a circle. We know a circle of radius π
centered at the point β, π will have the equation π₯ minus β all squared plus π¦
minus π all squared is equal to π squared. And in this case, weβre already
told the center of our circle, and weβre already told the radius of our circle. Our circle is centered at the point
four, negative seven. So the value of β is four, and the
value of π is negative seven. And our circle has a radius of 10,
so our value of π is equal to 10.

So we just need to substitute these
values into our equation for a circle. Substituting in the center of our
circle and the radius of our circle, we get the equation π₯ minus four all squared
plus π¦ minus negative seven all squared is equal to 10 squared. Of course, this is not in the
general form of the equation of a circle. To do this, we need to distribute
our squares over our parentheses. Thereβs a few different ways of
doing this. For example, we could use binomial
expansion or the FOIL method. Either way, distributing the square
over our first set of parentheses, we get π₯ squared minus eight π₯ plus 16.

To distribute the square over our
second set of parentheses, it might be easier to rewrite this as π¦ plus seven all
squared. Then, if we distribute this, we get
π¦ squared plus 14π¦ plus 49. The last thing weβll do is evaluate
10 squared to give us 100. Finally, all we need to do is
rewrite this in the form given to us in the question. To start, we rearrange these four
terms to give us π₯ squared plus π¦ squared minus eight π₯ plus 14π¦. Then, we want to subtract 100 from
both sides of the equation. Doing this, we can see we get a
constant term of 16 plus 49 minus 100. And we can calculate this; itβs
equal to negative 35.

So we get the equation π₯ squared
plus π¦ squared minus eight π₯ plus 14π¦ minus 35 is equal to zero. And we can now see this is in the
general form given to us in the question. Therefore, we were able to show the
general form of the equation of a circle of radius 10 centered at the point four,
negative seven is given by the equation π₯ squared plus π¦ squared minus eight π₯
plus 14π¦ minus 35 is equal to zero.

Letβs now look at an example about
finding the equation of a circle, given its center and a point which lies on the
circle.

Determine the equation of a circle
that passes through the point π΄: zero, eight if its center is π: negative two,
negative six.

The question wants us to find the
equation of a circle. Weβre told that this circle passes
through the point π΄ which has coordinates zero, eight. And the center of this circle is
the point π which has coordinates negative two, negative six. Letβs start by sketching what we
know about this circle. To start, weβll plot the center of
our circle on our graph. Thatβs the point π. It has coordinates negative two,
negative six. Next, weβll also plot the point π΄
on our graph. Remember, the circle passes through
this point and it has coordinates zero, eight. We donβt actually need to sketch
our circle to answer this question; however, weβll sketch a segment of this circle
to make it easier to visualize.

Letβs now recall what we know about
the equations for circles. We know that a circle centered at
the point β, π with a radius of π will have the equation π₯ minus β all squared
plus π¦ minus π all squared is equal to π squared. In other words, to find the
equation of a circle, we just need to know the coordinates of its center and its
radius. Of course, in this case, weβre
already told the coordinates of the center of the circle. Weβre told its center is negative
two, negative six. So we can set our value of β equal
to negative two and our value of π equal to negative six. This means to find the equation of
this circle, all we need to do is find its radius.

Remember, π will be the length
from the center of our circle to any point on our circle. We only know the coordinates of one
point on our circle. Thatβs the point π΄. So in our case, to find the value
of π, we need to find the length of the line between negative two, negative six and
zero, eight. And thereβs a few different ways of
approaching this problem. Weβre going to solve this by
drawing the following right-angled triangle. Weβre going to connect the point π
to the π¦-axis by using a horizontal line. The width of this triangle will
then be the absolute value of the π₯-coordinate of π, which is two. The height of this section will be
the absolute value of the π¦-coordinate of π, which is six. And the height of this section of
the triangle will be the π¦-coordinate of π΄, which is eight.

Then, we can just combine these two
lengths together to find that the height of our triangle will be eight plus six,
which is 14. Now, we have a right-angled
triangle where we know its width and we know its height. So we can apply the Pythagorean
theorem to find the length of its hypotenuse. This tells us π squared is equal
to two squared plus 14 squared. If we calculate two squared plus 14
squared, we get 200. And at this point, we can calculate
the radius. It would be the positive square
root of 200 because, remember, the radius is a length, so it must be positive. However, in our formula, we only
need the value of π squared, which we know is 200.

So we can just substitute π
squared is equal to 200, β is equal to negative two, and π is equal to negative six
into our equation for a circle. This gives us π₯ minus negative two
all squared plus π¦ minus negative six all squared is equal to 200. Finally, we can simplify π₯ minus
negative two to give us π₯ plus two and π¦ minus negative six to give us π¦ plus
six. And this gives us our final
answer. Therefore, we were able to show
given that a circle has its center at the point π, which is negative two, negative
six, and the circle also passes through the point π΄, which has coordinates zero,
eight, we were able to show the equation of this circle must be π₯ plus two all
squared plus π¦ plus six all squared is equal to 200.

Letβs now go over the key points of
this video. First, by using the definition of a
circle and the Pythagorean theorem, we were able to derive that a circle of radius
π with its center at the point β, π will have the equation π₯ minus β all squared
plus π¦ minus π all squared is equal to π squared. We also saw that we can find the
center and radius of a circle just from its equation. If a circle has the equation π₯
minus β all squared plus π¦ minus π all squared plus π squared, then its center
will be the point β, π and its radius will be π, of course, where π is a positive
number because it represents a length.

We also showed that we can find the
equation of a circle given only the coordinates of its center and a point on the
circle. And this is because we can find the
radius in this case because itβs just the distance between the center and the point
that weβre given. And we could find this distance by
using the Pythagorean theorem. Finally, by distributing the
squares over the parentheses in our equation for a circle, we were able to show that
the equation of a circle in its general form is given by the equation π₯ squared
plus π¦ squared plus ππ₯ plus ππ¦ plus π is equal to zero for some constants π,
π, and π.