### Video Transcript

In this video, we will learn how to
find the power of a point with respect to a circle and use the power of a point to
find other geometric lengths. The power of a point is a number
which quantifies the geometric relationship between a point and a circle. We define the power of a point as
follows. Given a circle of radius π
centered at π and a point π΄, the power of the point π΄ with respect to the circle
π, denoted P π of π΄, is given by P π of π΄ is equal to π΄π squared minus π
squared. Letβs now look at an example of how
we can use this definition of the power of a point in order to find the power of a
point with respect to a given circle.

A circle has center π and radius
π is equal to 21. Find the power of the point π΄ with
respect to the circle given that π΄π is equal to 25.

In order to answer this question,
weβre going to need the definition of the power of a point. We have that the power of the point
π΄ with respect to the circle π or P π of π΄ is equal to π΄π squared minus π
squared. Now, in the question, weβve been
given that π΄π is equal to 25. Weβve also been given that π is
equal to 21. All we need to do is substitute
these values into our formula for the power of the point π΄. We have that P π of π΄ is equal to
25 squared minus 21 squared.

Now, 25 squared is equal to 625 and
21 squared is equal to 441. And the difference between these
numbers is 184. Here, we have reached our solution,
which is that the power of point π΄ with respect to the circle π is equal to
184. In this example, our power of the
point π΄ with respect to the circle π is positive, and this happens because π΄π is
larger than the radius π. This in fact tells us that the
point π΄ is outside of the circle.

There are two other cases to
consider regarding the position of point π΄ in relation to the circle π. Shown here is the first case where
π΄ is outside of the circle, much like in the previous example. We can see that the length of π΄π
is greater than π; hence, π΄π squared is greater than π squared. Subtracting π squared from both
sides of the equation, we can see that π΄π squared minus π squared is greater than
zero. Now, the left-hand side of this
inequality is equal to the power of the point π΄ with respect to the circle π. Hence, P π of π΄ is greater than
zero, which is equivalent to say that it is positive.

The second case is when π΄ is on
the circle. As we can see from the diagram,
π΄π is equal to the length of the radius. Hence, π΄π is equal to π. Following similar logic to the
first case, we have that π΄π squared is equal to π squared and π΄π squared minus
π squared is equal to zero. Hence, when π΄ is on the
circumference of the circle, the power of the point π΄ with respect to the circle π
is equal to zero. The third case is when π΄ is inside
the circle. Hence, π΄π is less than π. Following similar logic to the
previous two cases, we obtain that π΄π squared minus π squared is less than
zero. Hence, when our point π΄ is inside
the circle, the power of the point is negative. Letβs look at an example of how
this can be used in order to determine the position of a point with respect to a
circle.

Determine the position of a point
π΄ with respect to the circle π if P π of π΄ is equal to 814.

We know that the power of the point
π΄ with respect to the circle π is equal to π΄π squared minus π squared. And we have been told that this
value is equal to 814 which is greater than zero. So P π of π΄ must be positive. This tells us that π΄π squared
minus π squared is positive too. Adding π squared to both sides of
the inequality, we obtain that π΄π squared is greater than π squared. Since π΄π and π are both lengths,
we know that they must both be greater than zero. Hence, when taking square roots on
both sides of our inequality, we donβt need to worry about any negatives, giving us
that π΄π is greater than π. Since the length from the center of
the circle to the point π΄ or π΄π is larger than the radius of the circle, hence,
this tells us our solution, which is that the point π΄ lies outside of the
circle.

Letβs now look at some more
geometric applications of the power of a point. We will start by considering the
relationship between the power of a point and the length of a tangent. In the diagram, we can see we have
a circle with center π, a point π΅ on the circumference, and a point π΄ such that
π΄π΅ is a tangent to the circle. Since we have defined π΄π΅ as a
tangent, we know that the measure of the angle ππ΅π΄ must be 90 degrees. Hence, the triangle ππ΅π΄ is a
right triangle. This means that we can apply the
Pythagorean theorem. We have that the square of the
hypotenuse or π΄π squared is equal to the sum of the square of the other two sides,
so π΄π΅ squared plus π squared.

Rearranging this equation, we have
that π΄π squared minus π squared is equal to π΄π΅ squared. Now the left-hand side of this
equation is equal to the power of the point π΄ with respect to the circle π. So we can say that P π of π΄ is
equal to π΄π΅ squared. This leads us to another definition
for the power of a point and the length of a tangent. Consider a circle π and a point π΄
outside the circle. Let π΄π΅ be a tangent to the
circle. Then P π of π΄ is equal to π΄π΅
squared. Next, we will consider a circle
containing a secant and a tangent. This property relating the length
of tangents and the power of a point will be helpful in our proof.

Here, we have a circle where π΄π΅
is a tangent to the circle, π΄πΆπ· is a secant, and π΅πΆ and π΅π· are both chords
within the circle. Now, weβre going to claim that
triangle π΄π΅πΆ is similar to triangle π΄π΅π·. And weβre going to try and prove
this claim. To start off with, we can see that
the angle π΅π΄πΆ is shared by both triangles. Hence, all we need to do is show
that the triangles share one more pair of equal angles. Letβs start by calling the measure
of the arc π΅πΆ π. So this central angle subtended by
this arc will also be equal to π. Now, we know that the inscribed
angle which subtends an arc is equal to half the central angle subtending the same
arc.

Hence, we can say the angle π΅π·πΆ
is equal to one-half of π. We also know that the angle between
a tangent and a chord is half the measure of the arc incepted by the chord. So we have that the angle π΄π΅πΆ is
also equal to one-half π. So we have found two pairs of equal
angles within our triangles, thus proving our claim that the two triangles are
similar. Using the similarity of these two
triangles, we can form a relationship between the corresponding side lengths. We have that π΄π΅ over π΄π· is
equal to π΄πΆ over π΄π΅. Multiplying through by the
denominators, we obtain that π΄π΅ squared is equal to π΄πΆ multiplied by π΄π·.

Now π΄π΅ is a tangent to the
circle. And we know that when π΄π΅ is a
tangent, the power of the point π΄ with respect to the circle π is equal to π΄π΅
squared. Equating this definition with the
equation we have just found, we can say that P π of π΄ is equal to π΄πΆ multiplied
by π΄π·. We have arrived at another
definition, this one for the power of a point and the length of a secant. Consider a circle π and a point π΄
outside the circle. Let π΄πΆπ· be a secant to the
circle. Then P π of π΄ is equal to π΄πΆ
times π΄π·. When the point π΄ is outside of our
circle, the power of π΄ with respect to the circle forms a relationship with both a
tangent to the circle and a secant to the circle.

Using this, weβre able to form a
relationship between tangents and secants, both passing through the same point. This is known as the power of a
point theorem. The power of a point theorem
consists of three different statements, all relating to the power of a point. Letβs start by looking at the power
of a point theorem relating a tangent and a secant to the circle. Consider a circle π and a point π΄
outside the circle. Let π΄π΅ be a tangent to the circle
and π΄πΆπ· be a secant to the circle. Then we can say that π΄π΅ squared
is equal to π΄πΆ multiplied by π΄π·. Letβs now consider an example where
weβll use this theorem in order to find a missing length involving a tangent and a
secant to a circle.

A circle has a tangent π΄π΅ and a
secant π΄π· that cuts the circle at πΆ. Given that π΄π΅ is equal to seven
centimeters and π΄πΆ is equal to five centimeters, find the length of πΆπ·. Give your answer to the nearest
hundredth.

The first thing we notice about
this question is that weβre asked about a tangent and a secant, both of which pass
through the point π΄. Hence, we can use the power of a
point theorem for a tangent and a secant. This theorem tells us that for a
tangent π΄π΅ and a secant π΄πΆπ·, much like we have in this example, π΄π΅ squared is
equal to π΄πΆ multiplied by π΄π·. We have been told in the question
that π΄π΅ is equal to seven centimeters and π΄πΆ is equal to five centimeters. Substituting these values into our
formula, we obtain that seven squared is equal to five multiplied by π΄π·.

Rearranging, we obtain that π΄π· is
equal to seven squared over five or 49 over five. Writing this fraction as a number,
we obtain that the length of π΄π· is 9.8 centimeters. We may have found π΄π·, but this is
not quite our solution yet, since the question asked us to find the length of
πΆπ·. We can use the fact that the secant
π΄πΆπ· is in fact a line segment. So π΄π· will be equal to π΄πΆ plus
πΆπ·. We have just found π΄π·, and weβve
been given π΄πΆ in the question. So we have that 9.8 is equal to
five plus πΆπ·. Rearranging this, we obtain that
πΆπ· is equal to 4.8. We mustnβt forget that the question
asked us to give our answer to the nearest hundredth and also our units of
centimeters. Therefore, our solution is that
πΆπ· is equal to 4.80 centimeters.

We have seen how we can solve a
problem involving a tangent and a secant. Next, weβll consider what happens
when we have two secants. Here, we have a circle with two
secants, π΄π΅πΆ and π΄π·πΈ. The general property of the power
of a point for secants tells us that for secant π΄πΆπ·, P π of π΄ is equal to π΄πΆ
times π΄π·. In our diagram, we have the secants
π΄π΅πΆ and π΄π·πΈ. So applying this property to the
two secants, we have that P π of π΄ is equal to π΄π΅ times π΄πΆ and also that P π
of π΄ is equal to π΄π· times π΄πΈ. By equating the right-hand sides of
these two equations, we will arrive at our secant power of a point theorem.

Consider a circle π and a point π΄
outside the circle. Let π΄π΅πΆ and π΄π·πΈ be secants to
the circle. Then π΄π΅ times π΄πΆ is equal to
π΄π· times π΄πΈ. Weβll now look at an example of how
this theorem can be used.

A circle has two secants π΄π΅ and
π΄π·, intersecting at π΄. Given that π΄πΈ is equal to three
centimeters, πΈπ· is equal to five centimeters, and π΄π΅ is equal to nine
centimeters, find the length of π΅πΆ, giving your answer to the nearest tenth.

We can start by noticing that we
have two secants to the circle intersecting outside at the point π΄. Hence, the power of a point theorem
tells us that π΄πΆ times π΄π΅ is equal to π΄πΈ times π΄π·. In the question, weβve been given
the length of π΄πΈ, πΈπ·, and π΄π΅. Since π΄πΈπ· is a line segment, so
π΄π· is equal to π΄πΈ plus πΈπ·. We can substitute in the values and
simplify to find that π΄π· is equal to eight centimeters. Now, we can substitute π΄π΅, π΄πΈ,
and π΄π· into our equation from the power of a point theorem. Next, we divide both sides by nine
and cancel a factor of three. This gives us that π΄πΆ is equal to
eight over three centimeters.

In order to find the length π΅πΆ,
we can use the fact that π΄πΆπ΅ is a line segment to say that π΄π΅ is equal to π΄πΆ
plus π΅πΆ. So nine is equal to eight over
three plus π΅πΆ. Rearranging and simplifying, we
find that π΅πΆ is equal to 19 over three centimeters. Rounding this answer to the nearest
tenth, we reach our solution, which is that π΅πΆ is equal to 6.3 centimeters.

For the final power of a point
theorem, weβll be considering a circle containing two chords. Letβs start by supposing that one
of those chords is in fact a diameter. Here, we have a chord π΅πΆ
intersecting our diameter π·πΈ. Weβre going to make a claim that
triangle π΄π΅π· is similar to triangle π΄πΆπΈ. Immediately, we can see that angles
πΆπ΄πΈ and π·π΄π΅ are opposite angles. Hence, theyβre equal. All we need to do to prove
similarity is show that two other angles in the triangles are equal. If we consider the arc π·πΆ, we can
see that the two angles π·π΅πΆ and π·πΈπΆ are inscribed by this arc. Hence, they are equal. So we have proven our claim that
these two triangles are similar.

Using this similarity, we can form
a relationship between the corresponding side lengths. π΄π΅ over π·π΄ is equal to π΄πΈ
over π΄πΆ. Multiplying through by the
denominators, we have that π΄π΅ times π΄πΆ is equal to π΄πΈ times π·π΄. We can now link this equation to
the power of a point. Since our point is inside the
circle, we know that the power of the point will be negative. So in order to keep our values
positive, letβs multiply both sides of this equation by negative one. Now, we can factorize the
right-hand side of the equation. From our diagram, we can see that
π·π is equal to π. We can also see that π·π΄ is equal
to π·π minus π΄π, which is also equal to π minus π΄π. Similarly, ππΈ is equal to π. And π΄πΈ is equal to ππΈ plus
π΄π, which is also equal to π plus π΄π.

Since π·π΄ is equal to π minus
π΄π and π΄πΈ is equal to π plus π΄π, we can say that negative P π of π΄ is equal
to π·π΄ times π΄πΈ. Hence, we can relate the power of a
point to the chord π΅π΄πΆ. We can say that when π΄ is inside
the circle, negative P π of π΄ is equal to π΄π΅ times π΄πΆ. Now, letβs consider a circle
containing two chords. Here we have the chords π΅π΄πΆ and
π·π΄πΈ. For the chord π΅π΄πΆ, we can say
that negative P π of π΄ is equal to π΄π΅ times π΄πΆ. And for the chord π·π΄πΈ, we can
say that negative P π of π΄ is equal to π΄π· times π΄πΈ. By equating these two equations, we
reach our third power of a point theorem.

Consider a circle π and a point π΄
inside the circle. Let π΅π΄πΆ and π·π΄πΈ be chords to
the circle. Then π΄π΅ times π΄πΆ is equal to
π΄π· times π΄πΈ. Letβs now look at an example of how
this theorem can be used.

A circle has two chords, π΄πΆ and
π΅π·, intersecting at πΈ. Given that π΄πΈ to π΅πΈ is equal to
one to three and πΆπΈ is equal to six centimeters, find the length of π·πΈ.

The power of a point theorem for
two chords tells us that π΄πΈ times πΆπΈ is equal to π·πΈ times π΅πΈ. Now weβve been given that πΆπΈ is
equal to six and also that π΄πΈ to π΅πΈ is equal to one to three. If we let π΄πΈ be equal to π₯
centimeters, then we can say that π΅πΈ is equal to three π₯ centimeters. Now, we can substitute our values
into our formula. We have that π₯ times six is equal
to π·πΈ times three π₯. Rearranging this for π·πΈ, we find
that π·πΈ is equal to two centimeters.

Weβve now covered a variety of
examples. Letβs recap some key points of the
video.

Key Points

Given a circle of radius π
centered at π and a point π΄, the power of point π΄ with respect to the circle is
given by P π of π΄ is equal to π΄π squared minus π squared. If P π of π΄ is greater than zero,
then π΄ lies outside the circle. If P π of π΄ is equal to zero,
then π΄ lies on the circle. And if P π of π΄ is less than
zero, then π΄ lies inside the circle. The power of a point theorem which
is in three parts. For a tangent and a secant, π΄π΅
squared is equal to π΄πΆ times π΄π·. For two secants, π΄π΅ times π΄πΆ is
equal to π΄π· times π΄πΈ. And finally, for two chords, π΄π΅
times π΄πΆ is equal to π΄π· times π΄πΈ.