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 𝑟
centered at 𝑀 and a point 𝐴, the power of the point 𝐴 with respect to the circle
𝑀, denoted P 𝑀 of 𝐴, is given by P 𝑀 of 𝐴 is equal to 𝐴𝑀 squared minus 𝑟
squared. Let’s 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 𝑀 and radius
𝑟 is equal to 21. Find the power of the point 𝐴 with
respect to the circle given that 𝐴𝑀 is equal to 25.
In order to answer this question,
we’re going to need the definition of the power of a point. We have that the power of the point
𝐴 with respect to the circle 𝑀 or P 𝑀 of 𝐴 is equal to 𝐴𝑀 squared minus 𝑟
squared. Now, in the question, we’ve been
given that 𝐴𝑀 is equal to 25. We’ve also been given that 𝑟 is
equal to 21. All we need to do is substitute
these values into our formula for the power of the point 𝐴. We have that P 𝑀 of 𝐴 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 𝐴 with respect to the circle 𝑀 is equal to
184. In this example, our power of the
point 𝐴 with respect to the circle 𝑀 is positive, and this happens because 𝐴𝑀 is
larger than the radius 𝑟. This in fact tells us that the
point 𝐴 is outside of the circle.
There are two other cases to
consider regarding the position of point 𝐴 in relation to the circle 𝑀. Shown here is the first case where
𝐴 is outside of the circle, much like in the previous example. We can see that the length of 𝐴𝑀
is greater than 𝑟; hence, 𝐴𝑀 squared is greater than 𝑟 squared. Subtracting 𝑟 squared from both
sides of the equation, we can see that 𝐴𝑀 squared minus 𝑟 squared is greater than
zero. Now, the left-hand side of this
inequality is equal to the power of the point 𝐴 with respect to the circle 𝑀. Hence, P 𝑀 of 𝐴 is greater than
zero, which is equivalent to say that it is positive.
The second case is when 𝐴 is on
the circle. As we can see from the diagram,
𝐴𝑀 is equal to the length of the radius. Hence, 𝐴𝑀 is equal to 𝑟. Following similar logic to the
first case, we have that 𝐴𝑀 squared is equal to 𝑟 squared and 𝐴𝑀 squared minus
𝑟 squared is equal to zero. Hence, when 𝐴 is on the
circumference of the circle, the power of the point 𝐴 with respect to the circle 𝑀
is equal to zero. The third case is when 𝐴 is inside
the circle. Hence, 𝐴𝑀 is less than 𝑟. Following similar logic to the
previous two cases, we obtain that 𝐴𝑀 squared minus 𝑟 squared is less than
zero. Hence, when our point 𝐴 is inside
the circle, the power of the point is negative. Let’s 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
𝐴 with respect to the circle 𝑁 if P 𝑁 of 𝐴 is equal to 814.
We know that the power of the point
𝐴 with respect to the circle 𝑁 is equal to 𝐴𝑁 squared minus 𝑟 squared. And we have been told that this
value is equal to 814 which is greater than zero. So P 𝑁 of 𝐴 must be positive. This tells us that 𝐴𝑁 squared
minus 𝑟 squared is positive too. Adding 𝑟 squared to both sides of
the inequality, we obtain that 𝐴𝑁 squared is greater than 𝑟 squared. Since 𝐴𝑁 and 𝑟 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 don’t need to worry about any negatives, giving us
that 𝐴𝑁 is greater than 𝑟. Since the length from the center of
the circle to the point 𝐴 or 𝐴𝑁 is larger than the radius of the circle, hence,
this tells us our solution, which is that the point 𝐴 lies outside of the
circle.
Let’s 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 𝑀, a point 𝐵 on the circumference, and a point 𝐴 such that
𝐴𝐵 is a tangent to the circle. Since we have defined 𝐴𝐵 as a
tangent, we know that the measure of the angle 𝑀𝐵𝐴 must be 90 degrees. Hence, the triangle 𝑀𝐵𝐴 is a
right triangle. This means that we can apply the
Pythagorean theorem. We have that the square of the
hypotenuse or 𝐴𝑀 squared is equal to the sum of the square of the other two sides,
so 𝐴𝐵 squared plus 𝑟 squared.
Rearranging this equation, we have
that 𝐴𝑀 squared minus 𝑟 squared is equal to 𝐴𝐵 squared. Now the left-hand side of this
equation is equal to the power of the point 𝐴 with respect to the circle 𝑀. So we can say that P 𝑀 of 𝐴 is
equal to 𝐴𝐵 squared. This leads us to another definition
for the power of a point and the length of a tangent. Consider a circle 𝑀 and a point 𝐴
outside the circle. Let 𝐴𝐵 be a tangent to the
circle. Then P 𝑀 of 𝐴 is equal to 𝐴𝐵
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 𝐴𝐵
is a tangent to the circle, 𝐴𝐶𝐷 is a secant, and 𝐵𝐶 and 𝐵𝐷 are both chords
within the circle. Now, we’re going to claim that
triangle 𝐴𝐵𝐶 is similar to triangle 𝐴𝐵𝐷. And we’re going to try and prove
this claim. To start off with, we can see that
the angle 𝐵𝐴𝐶 is shared by both triangles. Hence, all we need to do is show
that the triangles share one more pair of equal angles. Let’s start by calling the measure
of the arc 𝐵𝐶 𝜃. So this central angle subtended by
this arc will also be equal to 𝜃. 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 𝐵𝐷𝐶
is equal to one-half of 𝜃. 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 𝐴𝐵𝐶 is
also equal to one-half 𝜃. 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 𝐴𝐵 over 𝐴𝐷 is
equal to 𝐴𝐶 over 𝐴𝐵. Multiplying through by the
denominators, we obtain that 𝐴𝐵 squared is equal to 𝐴𝐶 multiplied by 𝐴𝐷.
Now 𝐴𝐵 is a tangent to the
circle. And we know that when 𝐴𝐵 is a
tangent, the power of the point 𝐴 with respect to the circle 𝑀 is equal to 𝐴𝐵
squared. Equating this definition with the
equation we have just found, we can say that P 𝑀 of 𝐴 is equal to 𝐴𝐶 multiplied
by 𝐴𝐷. We have arrived at another
definition, this one for the power of a point and the length of a secant. Consider a circle 𝑀 and a point 𝐴
outside the circle. Let 𝐴𝐶𝐷 be a secant to the
circle. Then P 𝑀 of 𝐴 is equal to 𝐴𝐶
times 𝐴𝐷. When the point 𝐴 is outside of our
circle, the power of 𝐴 with respect to the circle forms a relationship with both a
tangent to the circle and a secant to the circle.
Using this, we’re 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. Let’s start by looking at the power
of a point theorem relating a tangent and a secant to the circle. Consider a circle 𝑀 and a point 𝐴
outside the circle. Let 𝐴𝐵 be a tangent to the circle
and 𝐴𝐶𝐷 be a secant to the circle. Then we can say that 𝐴𝐵 squared
is equal to 𝐴𝐶 multiplied by 𝐴𝐷. Let’s now consider an example where
we’ll use this theorem in order to find a missing length involving a tangent and a
secant to a circle.
A circle has a tangent 𝐴𝐵 and a
secant 𝐴𝐷 that cuts the circle at 𝐶. Given that 𝐴𝐵 is equal to seven
centimeters and 𝐴𝐶 is equal to five centimeters, find the length of 𝐶𝐷. Give your answer to the nearest
hundredth.
The first thing we notice about
this question is that we’re asked about a tangent and a secant, both of which pass
through the point 𝐴. Hence, we can use the power of a
point theorem for a tangent and a secant. This theorem tells us that for a
tangent 𝐴𝐵 and a secant 𝐴𝐶𝐷, much like we have in this example, 𝐴𝐵 squared is
equal to 𝐴𝐶 multiplied by 𝐴𝐷. We have been told in the question
that 𝐴𝐵 is equal to seven centimeters and 𝐴𝐶 is equal to five centimeters. Substituting these values into our
formula, we obtain that seven squared is equal to five multiplied by 𝐴𝐷.
Rearranging, we obtain that 𝐴𝐷 is
equal to seven squared over five or 49 over five. Writing this fraction as a number,
we obtain that the length of 𝐴𝐷 is 9.8 centimeters. We may have found 𝐴𝐷, but this is
not quite our solution yet, since the question asked us to find the length of
𝐶𝐷. We can use the fact that the secant
𝐴𝐶𝐷 is in fact a line segment. So 𝐴𝐷 will be equal to 𝐴𝐶 plus
𝐶𝐷. We have just found 𝐴𝐷, and we’ve
been given 𝐴𝐶 in the question. So we have that 9.8 is equal to
five plus 𝐶𝐷. Rearranging this, we obtain that
𝐶𝐷 is equal to 4.8. We mustn’t 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
𝐶𝐷 is equal to 4.80 centimeters.
We have seen how we can solve a
problem involving a tangent and a secant. Next, we’ll consider what happens
when we have two secants. Here, we have a circle with two
secants, 𝐴𝐵𝐶 and 𝐴𝐷𝐸. The general property of the power
of a point for secants tells us that for secant 𝐴𝐶𝐷, P 𝑀 of 𝐴 is equal to 𝐴𝐶
times 𝐴𝐷. In our diagram, we have the secants
𝐴𝐵𝐶 and 𝐴𝐷𝐸. So applying this property to the
two secants, we have that P 𝑀 of 𝐴 is equal to 𝐴𝐵 times 𝐴𝐶 and also that P 𝑀
of 𝐴 is equal to 𝐴𝐷 times 𝐴𝐸. By equating the right-hand sides of
these two equations, we will arrive at our secant power of a point theorem.
Consider a circle 𝑀 and a point 𝐴
outside the circle. Let 𝐴𝐵𝐶 and 𝐴𝐷𝐸 be secants to
the circle. Then 𝐴𝐵 times 𝐴𝐶 is equal to
𝐴𝐷 times 𝐴𝐸. We’ll now look at an example of how
this theorem can be used.
A circle has two secants 𝐴𝐵 and
𝐴𝐷, intersecting at 𝐴. Given that 𝐴𝐸 is equal to three
centimeters, 𝐸𝐷 is equal to five centimeters, and 𝐴𝐵 is equal to nine
centimeters, find the length of 𝐵𝐶, 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 𝐴. Hence, the power of a point theorem
tells us that 𝐴𝐶 times 𝐴𝐵 is equal to 𝐴𝐸 times 𝐴𝐷. In the question, we’ve been given
the length of 𝐴𝐸, 𝐸𝐷, and 𝐴𝐵. Since 𝐴𝐸𝐷 is a line segment, so
𝐴𝐷 is equal to 𝐴𝐸 plus 𝐸𝐷. We can substitute in the values and
simplify to find that 𝐴𝐷 is equal to eight centimeters. Now, we can substitute 𝐴𝐵, 𝐴𝐸,
and 𝐴𝐷 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 𝐴𝐶 is equal to
eight over three centimeters.
In order to find the length 𝐵𝐶,
we can use the fact that 𝐴𝐶𝐵 is a line segment to say that 𝐴𝐵 is equal to 𝐴𝐶
plus 𝐵𝐶. So nine is equal to eight over
three plus 𝐵𝐶. Rearranging and simplifying, we
find that 𝐵𝐶 is equal to 19 over three centimeters. Rounding this answer to the nearest
tenth, we reach our solution, which is that 𝐵𝐶 is equal to 6.3 centimeters.
For the final power of a point
theorem, we’ll be considering a circle containing two chords. Let’s start by supposing that one
of those chords is in fact a diameter. Here, we have a chord 𝐵𝐶
intersecting our diameter 𝐷𝐸. We’re going to make a claim that
triangle 𝐴𝐵𝐷 is similar to triangle 𝐴𝐶𝐸. Immediately, we can see that angles
𝐶𝐴𝐸 and 𝐷𝐴𝐵 are opposite angles. Hence, they’re 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 𝐷𝐶, we can
see that the two angles 𝐷𝐵𝐶 and 𝐷𝐸𝐶 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. 𝐴𝐵 over 𝐷𝐴 is equal to 𝐴𝐸
over 𝐴𝐶. Multiplying through by the
denominators, we have that 𝐴𝐵 times 𝐴𝐶 is equal to 𝐴𝐸 times 𝐷𝐴. 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, let’s 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
𝐷𝑀 is equal to 𝑟. We can also see that 𝐷𝐴 is equal
to 𝐷𝑀 minus 𝐴𝑀, which is also equal to 𝑟 minus 𝐴𝑀. Similarly, 𝑀𝐸 is equal to 𝑟. And 𝐴𝐸 is equal to 𝑀𝐸 plus
𝐴𝑀, which is also equal to 𝑟 plus 𝐴𝑀.
Since 𝐷𝐴 is equal to 𝑟 minus
𝐴𝑀 and 𝐴𝐸 is equal to 𝑟 plus 𝐴𝑀, we can say that negative P 𝑀 of 𝐴 is equal
to 𝐷𝐴 times 𝐴𝐸. Hence, we can relate the power of a
point to the chord 𝐵𝐴𝐶. We can say that when 𝐴 is inside
the circle, negative P 𝑀 of 𝐴 is equal to 𝐴𝐵 times 𝐴𝐶. Now, let’s consider a circle
containing two chords. Here we have the chords 𝐵𝐴𝐶 and
𝐷𝐴𝐸. For the chord 𝐵𝐴𝐶, we can say
that negative P 𝑀 of 𝐴 is equal to 𝐴𝐵 times 𝐴𝐶. And for the chord 𝐷𝐴𝐸, we can
say that negative P 𝑀 of 𝐴 is equal to 𝐴𝐷 times 𝐴𝐸. By equating these two equations, we
reach our third power of a point theorem.
Consider a circle 𝑀 and a point 𝐴
inside the circle. Let 𝐵𝐴𝐶 and 𝐷𝐴𝐸 be chords to
the circle. Then 𝐴𝐵 times 𝐴𝐶 is equal to
𝐴𝐷 times 𝐴𝐸. Let’s now look at an example of how
this theorem can be used.
A circle has two chords, 𝐴𝐶 and
𝐵𝐷, intersecting at 𝐸. Given that 𝐴𝐸 to 𝐵𝐸 is equal to
one to three and 𝐶𝐸 is equal to six centimeters, find the length of 𝐷𝐸.
The power of a point theorem for
two chords tells us that 𝐴𝐸 times 𝐶𝐸 is equal to 𝐷𝐸 times 𝐵𝐸. Now we’ve been given that 𝐶𝐸 is
equal to six and also that 𝐴𝐸 to 𝐵𝐸 is equal to one to three. If we let 𝐴𝐸 be equal to 𝑥
centimeters, then we can say that 𝐵𝐸 is equal to three 𝑥 centimeters. Now, we can substitute our values
into our formula. We have that 𝑥 times six is equal
to 𝐷𝐸 times three 𝑥. Rearranging this for 𝐷𝐸, we find
that 𝐷𝐸 is equal to two centimeters.
We’ve now covered a variety of
examples. Let’s recap some key points of the
video.
Key Points
Given a circle of radius 𝑟
centered at 𝑀 and a point 𝐴, the power of point 𝐴 with respect to the circle is
given by P 𝑀 of 𝐴 is equal to 𝐴𝑀 squared minus 𝑟 squared. If P 𝑀 of 𝐴 is greater than zero,
then 𝐴 lies outside the circle. If P 𝑀 of 𝐴 is equal to zero,
then 𝐴 lies on the circle. And if P 𝑀 of 𝐴 is less than
zero, then 𝐴 lies inside the circle. The power of a point theorem which
is in three parts. For a tangent and a secant, 𝐴𝐵
squared is equal to 𝐴𝐶 times 𝐴𝐷. For two secants, 𝐴𝐵 times 𝐴𝐶 is
equal to 𝐴𝐷 times 𝐴𝐸. And finally, for two chords, 𝐴𝐵
times 𝐴𝐶 is equal to 𝐴𝐷 times 𝐴𝐸.
