# Lesson Video: The Power of a Point Theorem Mathematics

In this video, we will learn how to find the power of a point with respect to a circle.

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### 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.