### Video Transcript

Constructing Circles

In this video, we will learn how to
construct circles given one, two, or three distinct points. We will also learn under which
conditions can we construct these circles and, in particular, how to find the
circumcircle of a triangle. Thatβs a circle which passes
through all three vertices of a triangle.

To begin, letβs start by recalling
what we mean by a circle. A circle is the set of all points a
specific distance from a point known as its center. In particular, the set distance is
usually known as the radius of the circle. So, every single point on our
circle will be equidistant from the point π. If we call the radius π, then the
distance between π and π΄ will be π for any point on our circle. And itβs also worth noting the line
segment between π and π΄ is called a radius of the circle. So the length of the radius is π,
and we also sometimes refer to this distance just as the radius as well.

Now that weβve recapped the
definition of a circle, weβre ready to start discussing the conditions under which
we can construct a circle. Letβs start by trying to construct
a circle through a given point π΄. We can do this by just choosing any
distinct point on the plane to be the center of our circle. Letβs label this point π. Then, since every point on our
circle needs to be equidistant from the point π, we can find the radius of our
circle. It will be the distance between π
and π΄. This then defines our circle
mathematically; it has center π and radius π.

However, we can also sketch this
circle by using a compass. We put the needle point of our
compass at the point π, and we adjust our compass so that the radius is going to be
equal to π, so our pencil lies at the point π΄. Moving the compass around then
sketches a circle which passes through the point π΄. But remember, we could choose any
distinct point in the plane π to be the center of our circle. This means that thereβs infinitely
many circles which pass through the point π΄. In fact, an even stronger result
holds true. Thereβs infinitely many circles
congruent to the circle we just drew which passes through the point π΄. Remember, for two shapes to be
congruent, they need to have the same size and shape. So for a circle to be congruent to
our given circle, it just needs to be a circle with the same radius.

To see why this is true, letβs
start by sketching a circle of radius π centered at the point π΄. Of course, since π is a distance
of π away from the point π΄, π lies on this circle. In fact, any point on this circle
is a distance of π away from the point π΄. For example, the distance between
the point π two and π΄ is π. But now we can sketch a circle of
radius π around the point π two. This then gives us a second circle
which passes through the point π΄, which also has radius π. Therefore, we sketched two circles
of radius π which both pass through the point π΄. These are congruent circles. And we can choose any point on the
circle of radius π around the point π΄ to be the center of our circle. This means there is infinitely many
congruent circles which pass through the point π΄ which have radius π.

Now that weβve seen how to
construct a circle through one given point, letβs move on to trying to construct a
circle between two given distinct points. Letβs say the points π΄ and π΅. To construct a circle, weβve seen
we need to find two things. We need to determine its center and
its radius. Since both points π΄ and π΅ need to
be on our circle, they need to be the same distance away from the center of our
circle. So the first thing weβre going to
need to do is find a point equidistant from π΄ and π΅. And to do this, weβre going to
recall the following result. Every point which is equidistant
from two points π΄ and π΅ will lie on the perpendicular bisector of the line segment
π΄π΅.

Before we see why this might be
true, letβs start by constructing the perpendicular bisector of the line segment
π΄π΅. Letβs start by sketching the line
segment π΄π΅, and weβll call the midpoint of this line segment π. And we know that π΄π and ππ΅ have
the same length. And in fact, this is already enough
to answer our question. π is equidistant from the point π΄
and π΅, so we could choose this to be the center of our circle. Then all we would do is choose the
radius to be the distance between π΄ and π or the distance between π and π΅. However, this is only one circle
which passes through π΄ and π΅. Letβs see if we can find more.

We then draw a line perpendicular
to the line segment π΄π΅ which passes through π. This is the perpendicular bisector
of the line segment π΄π΅. Our property then tells us any
point on this line is equidistant from π΄ and π΅, and every point which is
equidistant is also on this line. And we can actually prove that this
is true. For example, letβs show that π two
is equidistant from π΄ and π΅. We have two right triangles,
triangle π΄ππ two and triangle π΅ππ two. We can find an expression for the
hypotenuse of either triangle by using the Pythagorean theorem.

Remember, this tells us the square
of the hypotenuse is equal to the sum of the squares of the two shorter sides in a
right triangle. So in our left-right triangle, we
have the length of π΄π two squared is equal to the length of π΄π squared plus the
length of ππ two squared. But remember, the length of π΄π is
equal to the length of ππ΅. So in our expression, we can
replace the length of π΄π with ππ΅. This then gives us the length of
ππ΅ squared plus the length of ππ two squared.

But then by the Pythagorean
theorem, this is actually just the square of the hypotenuse in our other right
triangle. So by the Pythagorean theorem, this
is the length of π΅π two squared. Then, all we need to do is bring
down the left-hand side of our equation. The length of π΄π two squared is
equal to the length of π΅π two squared. Then, since both of these are
lengths, we can just take the positive square root of both sides of the
equation. The distance between π΄ and π two
is the same as the distance between π΅ and π two. Therefore, any line on our circle
is equidistant from π΄ and π΅, which means it can be the center of a circle which
passes through both π΄ and π΅.

For example, to sketch a circle
from the point π three which passes through π΄ and π΅, we just put our compass
point at the point π three and then choose the radius to be the distance between π
three and either π΄ or π΅. Once again, because we can choose
any point on the perpendicular bisector, thereβs infinitely many circles which pass
through both π΄ and π΅. Letβs go for a question which will
allow us to find the smallest radius circle which passes through two points π΄ and
π΅.

Consider the two points π΄ and
π΅. What is the radius of the smallest
circle that can be drawn in order to pass through the two points?

In this question, we need to
determine the radius of the smallest circle which passes through two points π΄ and
π΅. To answer this question, we start
by recalling that every point on a circle will be equidistant from its center. And we also recall another
fact. Every point which is equidistant
from two points π΄ and π΅ will lie on the perpendicular bisector of π΄ and π΅. So the center of our circle must
lie on the perpendicular bisector of π΄ and π΅. So letβs start by constructing
this. We need to draw the line segment
from π΄ to π΅ and then mark the midpoint of this line segment. Weβll call this π. Then the perpendicular bisector is
the line perpendicular to π΄π΅ which passes through π. Any point which lies on this line
is equidistant from π΄ and π΅.

And so these can be choices for the
center of our circle. For example, π, π two, π three,
or π four could be the centers of our circle. Then the radius of this circle
would just be the distance between the center and any point on the circle, for
example, the distance between the center and point π΄. Graphically, it appears the further
we get away from the point π, the larger this radius is. In fact, we can prove this is
true. For example, we can see that
triangle π΄ππ two is a right triangle. In particular, this means its
hypotenuse must be longer than the other sides. π΄ππ three is also a right
triangle, and π΄ππ four is another right triangle. So π three is bigger than π, and
π four is bigger than π. Therefore, π will be the circle
with the smallest radius which passes through π΄ and π΅.

And although we were not asked to
in this question, we can even sketch this circle by putting the points of our
compass at the point π and then the tip of our compass at either the points π΄ or
π΅. We were asked to find the radius of
this circle. Remember, we chose π to be the
midpoint of the line segment π΄π΅. And this means the value of π is
going to be one-half the distance from π΄ to π΅. In other words, the radius of the
smallest circle which passes through two distinct points π΄ and π΅ is one-half the
distance from π΄ to π΅.

So far, weβve constructed circles
through one point and through two distinct points. Now letβs try to construct the
circle through three distinct points π΄, π΅, and πΆ. To do this, we need to find the
center of our circle which needs to be equidistant from all three points. If the center of the circle is
equidistant from all three points, then it must be equidistant from both π΄ and
π΅. And we can find the set of all
points equidistant from both π΄ and π΅ by finding the perpendicular bisector. So we start by finding the midpoint
of the line segment between π΄ and π΅. And then we draw the line through
the midpoint π one which is perpendicular to the line segment π΄π΅. Every point equidistant from both
π΄ and π΅ lies on the perpendicular bisector between π΄ and π΅. So if the center of our circle
exists, it must lie on this line.

But the exact same reasoning is
true for points πΆ and π΅. The center of our circle is
equidistant from πΆ and π΅, so it must lie on the perpendicular bisector of the line
segment π΅πΆ. So weβll also sketch the
perpendicular bisector of line segment π΅πΆ. We find the point π two which is
the midpoint of this line segment and then draw a line perpendicular to this line
which passes through π two. Every point on this line is
equidistant from π΅ and πΆ, and we can see thereβs a point of intersection between
the two perpendicular bisectors. Since this point is on both
perpendicular bisectors, itβs equidistant from π΄, π΅, and πΆ. This means we can construct a
circle with center π which passes through all three points π΄, π΅, and πΆ. The radius of this circle will then
be the distance between either π and π΄, π and π΅, or π and πΆ.

We can then ask another
question. Is there another circle which
passes through these three points? And in fact, using the exact same
reasoning, we can show that there is not. This is the only circle which
passes through these three points. To see why this is true, remember,
the center of our circle must lie on the perpendicular bisector between π΄ and π΅
and the perpendicular bisector between π΅ and πΆ. And in the plane, distinct lines
can only ever intersect in at most one point. So π is the only point on both of
these lines. Therefore, itβs the only possible
center of this circle. And so this is the unique circle
which passes through π΄, π΅, and πΆ.

In fact, thereβs slightly more we
can show. What if we wanted a circle which
also passes through another point π·? We can see that π· does not lie on
the circle. And since this is the only circle
which passes through π΄, π΅, and πΆ, we can conclude that there is not a circle
which passes through all four of these points. But weβve only shown that we can
construct a circle between these three points. What if we had three different
points? Can we do this in general? In our next example, weβll discuss
the condition under which we can sketch a circle through three points.

True or false: if a circle passes
through three points, then the three points should belong to the same line.

To answer this question, letβs
start by sketching a circle. We can choose three points on this
circle. Letβs label these π΄, π΅, and
πΆ. We can see that these three points
do not lie on the same straight line. However, the circle does pass
through these three points. This is enough to prove the
statement is false. If a circle passes through three
points, then the three points do not need to belong to the same line.

This could make us ask an
interesting question. What would happen if the three
points were in a straight line? For example, can we construct a
circle between these three points π, π, and π
? The center of the circle must be
equidistant from both three of these points. In particular, the center of the
circle must be equidistant from both π and π. And we know that the perpendicular
bisector of π and π contains all points equidistant from both π and π. So the center of our circle must
lie on this line. Similarly, the center of this
circle is equidistant from π and π
, so it must lie on the perpendicular bisector
of π and π
.

And then we see a problem. These two lines are parallel. They never intersect, so thereβs no
point equidistant from π, π, and π
. This leads to a slightly stronger
result. There is no circle which passes
through three points which lie on the same straight line. But we can answer this question as
false. A circle that passes through three
points do not need to have the three points lie on the same straight line. The statement is false.

In fact, thereβs an even stronger
result still that weβll need to know. If three points are noncolinear,
which means they donβt lie on the same straight line, then there exists a unique
circle which passes through all three of the points. And we saw how to construct this
circle earlier. We just find the intersection of
the perpendicular bisectors of two pairs of the points. This gives us the center of our
circle, and then we can find the radius of our circle as the distance between the
center and any of the three points. Letβs now see an example of how we
can use this property on a problem involving triangles.

True or false: a circle can be
drawn through the vertices of any triangle.

To answer this question, we might
start by drawing a triangle and seeing how we might construct a circle which passes
through all three vertices of the triangle. However, this is not necessary
since we can just recall a useful fact about circles. We know that any three noncolinear
points have a unique circle which passes through all three points, where we remember
noncolinear points means they donβt lie on the same straight line. And the vertices of a triangle
donβt lie on the same straight line. Thatβs what it means for a shape to
be a triangle. For example, if we had three points
which lied on a straight line and we tried to connect the vertices together, we
would just end up with a straight line.

Therefore, we can just apply this
property to show that the answer is true. A circle can be drawn through the
vertices of any triangle. This is called the circumcircle of
the triangle.

Letβs now see an example where we
determine how to construct the circumcircle of a given triangle.

In the following two figures, two
types of construction have been made on the same triangle π΄π΅πΆ. Which point will be the center of
the circle that passes through the triangleβs vertices?

In this question, weβre asked to
determine the center of a circle which passes through all three of the triangleβs
vertices. Thatβs the points π΄, π΅, and
πΆ. And to do this, we need to
remember, the center of a circle will be equidistant from all points on the
circumference of the circle. Therefore, we need to determine
which points in either of the two diagrams is equidistant from all three points π΄,
π΅, and πΆ. To do this, we know in the second
diagram weβre given the perpendicular bisectors of all three sides of the
triangle. We can see that these are the
perpendicular bisectors because they cut the sides of the triangle at right
angles. And they cut the sides of the
triangle in half represented by either the one, two, or three lines.

This is useful because every single
point on the perpendicular bisector between π΅ and πΆ will be equidistant from π΅
and πΆ. For example, the point πΉ is
equidistant from both π΅ and πΆ. This result is true for any
perpendicular bisector, so we can see that πΉ also lies on the perpendicular
bisector between π΄ and π΅. So πΉ is also equidistant from π΄
and π΅. This means itβs equidistant from
all three vertices of the triangle. Therefore, if we draw a circle
centered at πΉ of radius the distance between π΄ and πΉ, we would get a circle which
passes through all three vertices of the triangle.

The same cannot be true for the
point π; although itβs found by taking the bisectors of each of the angles of our
triangle, π will not necessarily be equidistant from all three vertices of the
triangle. Therefore, we can conclude the
point πΉ is the center of a circle which passes through all three vertices of the
triangle.

In our final example, weβll go for
a property involving the number of intersections possible in two distinct
circles.

True or false: two distinct circles
can intersect at more than two points.

To answer this question, letβs
start by thinking about how circles can intersect. Thereβs two different ways that
circles canβt intersect. For example, we could have one
circle inside another circle, or we could just have two circles next to each
other. Itβs also possible for two circles
to intersect at a single point, and in fact, thereβs two different ways this can
happen. We can have both circles next to
each other, or we can have one circle inside of another circle. Both of these only intersect at a
single point. Finally, itβs also possible for
circles to intersect at two distinct points, for example, the following shape which
looks like a Venn diagram.

But what if we wanted two circles
which intersect at three distinct points? Well, if the three points of
intersection were noncolinear, then we know thereβs a unique circle which passes
through all three of these points. This is called the circumcircle of
triangle π΄π΅πΆ, and since itβs unique, we canβt have two distinct circles in this
case. But weβre not done yet. We also need to consider the case
if the three points of intersection lied on the same straight line. But this is also not possible
because we recall there is no circle which passes through three points which lie on
the same straight line. Therefore, weβve shown the answer
is false. Two distinct circles cannot
intersect at more than two points.

Letβs now go over the key points of
this video. First, we showed that there are an
infinite number of circles passing through any single point π΄. We can then construct this circle
by choosing any point to be our center π, and then the radius of our circle will be
the distance between π΄ and π. Next, we saw that there are an
infinite number of circles passing through any two distinct points π΄ and π΅. We can then construct this circle
by choosing any point on the perpendicular bisector of the line segment π΄π΅ to be
the center of our circle. And the radius of this circle will
then be the distance between π΄ and π or the distance between π΅ and π.

Next, we also saw that there is a
unique circle passing through any three noncolinear distinct points π΄, π΅, and
πΆ. We can then construct this circle
by recalling the center of this circle will be the intersection between the
perpendicular bisectors of the line segments between any two pairs of points, for
example, the perpendicular bisectors of the line segment π΄π΅ and π΅πΆ. And the radius of this circle will
be the distance between π΄ and π, π΅ and π, or πΆ and π. And itβs also worth pointing out we
showed that the noncolinearity condition is necessary. If the three points lie on the same
line, then there is no circle which passes through the three points. Finally, we saw that two distinct
circles will intersect at at most two points.