Lesson Video: The Power of a Point Theorem | Nagwa Lesson Video: The Power of a Point Theorem | Nagwa

# Lesson Video: The Power of a Point Theorem Mathematics • First Year of Secondary School

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

17:55

### 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. Letβ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 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, weβre 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, weβve been given that π΄π is equal to 25. Weβve 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. Letβ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 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 donβt 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.

Letβ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 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 ππ΅π΄ must be 90 degrees. Hence, the triangle ππ΅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΄πΆπ· is a secant, and π΅πΆ and π΅π· are both chords within the circle. Now, weβre going to claim that triangle π΄π΅πΆ is similar to triangle π΄π΅π·. And weβre going to try and prove this claim. To start off with, we can see that the angle π΅π΄πΆ is shared by both triangles. Hence, all we need to do is show that the triangles share one more pair of equal angles. Letβs start by calling the measure of the arc π΅πΆ 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·πΆ 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΅πΆ 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΄πΆ over π΄π΅. Multiplying through by the denominators, we obtain that π΄π΅ squared is equal to π΄πΆ 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΄πΆ 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΄πΆπ· be a secant to the circle. Then P π of π΄ is equal to π΄πΆ 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, weβ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. Letβs 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΄πΆπ· be a secant to the circle. Then we can say that π΄π΅ squared is equal to π΄πΆ multiplied by π΄π·. Letβs now consider an example where weβ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 π΄π΅ and a secant π΄π· that cuts the circle at πΆ. Given that π΄π΅ is equal to seven centimeters and π΄πΆ is equal to five centimeters, find the length of πΆπ·. Give your answer to the nearest hundredth.

The first thing we notice about this question is that weβre 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΄πΆπ·, much like we have in this example, π΄π΅ squared is equal to π΄πΆ multiplied by π΄π·. We have been told in the question that π΄π΅ is equal to seven centimeters and π΄πΆ 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 πΆπ·. We can use the fact that the secant π΄πΆπ· is in fact a line segment. So π΄π· will be equal to π΄πΆ plus πΆπ·. We have just found π΄π·, and weβve been given π΄πΆ in the question. So we have that 9.8 is equal to five plus πΆπ·. Rearranging this, we obtain that πΆπ· is equal to 4.8. We mustnβ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 πΆπ· is equal to 4.80 centimeters.

We have seen how we can solve a problem involving a tangent and a secant. Next, weβll consider what happens when we have two secants. Here, we have a circle with two secants, π΄π΅πΆ and π΄π·πΈ. The general property of the power of a point for secants tells us that for secant π΄πΆπ·, P π of π΄ is equal to π΄πΆ times π΄π·. In our diagram, we have the secants π΄π΅πΆ and π΄π·πΈ. So applying this property to the two secants, we have that P π of π΄ is equal to π΄π΅ times π΄πΆ and also that P π of π΄ is equal to π΄π· times π΄πΈ. 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΅πΆ and π΄π·πΈ be secants to the circle. Then π΄π΅ times π΄πΆ is equal to π΄π· times π΄πΈ. Weβll 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΄πΈ is equal to three centimeters, πΈπ· is equal to five centimeters, and π΄π΅ is equal to nine centimeters, find the length of π΅πΆ, 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΄πΆ times π΄π΅ is equal to π΄πΈ times π΄π·. In the question, weβve been given the length of π΄πΈ, πΈπ·, and π΄π΅. Since π΄πΈπ· is a line segment, so π΄π· is equal to π΄πΈ plus πΈπ·. 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΄πΈ, 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΄πΆ is equal to eight over three centimeters.

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

For the final power of a point theorem, weβll be considering a circle containing two chords. Letβs start by supposing that one of those chords is in fact a diameter. Here, we have a chord π΅πΆ intersecting our diameter π·πΈ. Weβre going to make a claim that triangle π΄π΅π· is similar to triangle π΄πΆπΈ. Immediately, we can see that angles πΆπ΄πΈ and π·π΄π΅ are opposite angles. Hence, theyβ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 π·πΆ, we can see that the two angles π·π΅πΆ and π·πΈπΆ 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΄πΈ over π΄πΆ. Multiplying through by the denominators, we have that π΄π΅ times π΄πΆ is equal to π΄πΈ 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, letβ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 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, ππΈ is equal to π. And π΄πΈ is equal to ππΈ plus π΄π, which is also equal to π plus π΄π.

Since π·π΄ is equal to π minus π΄π and π΄πΈ is equal to π plus π΄π, we can say that negative P π of π΄ is equal to π·π΄ times π΄πΈ. Hence, we can relate the power of a point to the chord π΅π΄πΆ. We can say that when π΄ is inside the circle, negative P π of π΄ is equal to π΄π΅ times π΄πΆ. Now, letβs consider a circle containing two chords. Here we have the chords π΅π΄πΆ and π·π΄πΈ. For the chord π΅π΄πΆ, we can say that negative P π of π΄ is equal to π΄π΅ times π΄πΆ. And for the chord π·π΄πΈ, we can say that negative P π of π΄ is equal to π΄π· times π΄πΈ. 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΄πΆ and π·π΄πΈ be chords to the circle. Then π΄π΅ times π΄πΆ is equal to π΄π· times π΄πΈ. Letβs now look at an example of how this theorem can be used.

A circle has two chords, π΄πΆ and π΅π·, intersecting at πΈ. Given that π΄πΈ to π΅πΈ is equal to one to three and πΆπΈ is equal to six centimeters, find the length of π·πΈ.

The power of a point theorem for two chords tells us that π΄πΈ times πΆπΈ is equal to π·πΈ times π΅πΈ. Now weβve been given that πΆπΈ is equal to six and also that π΄πΈ to π΅πΈ is equal to one to three. If we let π΄πΈ be equal to π₯ centimeters, then we can say that π΅πΈ 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·πΈ times three π₯. Rearranging this for π·πΈ, we find that π·πΈ is equal to two centimeters.

Weβve now covered a variety of examples. Letβs 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΄πΆ times π΄π·. For two secants, π΄π΅ times π΄πΆ is equal to π΄π· times π΄πΈ. And finally, for two chords, π΄π΅ times π΄πΆ is equal to π΄π· times π΄πΈ.

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