### Video Transcript

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.