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Special Segments in a Circle

  • If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal. AB × BC = DB × BE
  • If two secant segments are drawn to a circle from an exterior point, then the product of the measure of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment
  • If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment

Special Segments in a Circle

BF = 12, CF = 8, EF = 10, find DF.
  • BF*CF = EF*DF
  • DF = [BF*CF/EF]
  • DF = [12*8/10] = 9.6
DF = 9.6

BF = 12, EF = 20, BC ⊥DE , find DF.
  • BC ⊥DE
  • CF = BF = 12
  • BF*CF = EF*DF
  • DF = [BF*CF/EF]
  • DF = [12*12/20] = 7.2
DF = 7.2

BF = 12, CF = 4, EF = 6, find DF.
  • BF*CF = EF*DF
  • DF = [BF*CF/EF]
  • DF = [12*4/6] = 8
DF = 8

BC = 11, CF = 4, EF = 6, find DF.
  • BF = BC + CF
  • BF = 11 + 4 = 15.
  • BF*CF = EF*DF
  • DF = [BF*CF/EF]
  • DF = [15*4/6] = 10
DF = 10

DF = 12, EF = 6, find BF.
  • BF2 = DF*EF
  • BF2 = 12*6 = 72
  • BF = 6√2
BF = 6√2

BF = 6, EF = 3, find DE.
  • BF2 = DF*EF
  • 62 = DF*3
  • DF = 12
  • DE = DF − EF
  • DE = 12 − 3 = 8
DE = 8
, find x.
  • BF*CF = EF*DF
  • CF = [EF*DF/BF]
  • CF = [8*5/10] = 4.
  • x − 2 = 4
x = 6.
EF = 6, r = 4, find BF.
  • DF = 2r + EF
  • DF = 2*4 + 6 = 14
  • BF2 = DF*EF
  • BF2 = 14*6 = 84
BF = 2√{21}

find x.
  • DF = 10 + 4 = 14
  • BF = x + 6
  • BF*CF = EF*DF
  • (x + 6)*6 = 4*14
  • x = [4*14/6] − 6
x = 3.33

AE = 5, AD = 9, find AB.
  • AC2 = AE*AD
  • AC2 = 5*9 = 45
  • AC = 3√5
AB = AC = 3√5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Special Segments in a Circle

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Chord Segments 0:05
    • Chord Segments
  • Secant Segments 1:36
    • Secant Segments
  • Tangent and Secant Segments 4:10
    • Tangent and Secant Segments
  • Extra Example 1: Special Segments in a Circle 5:53
  • Extra Example 2: Special Segments in a Circle 7:58
  • Extra Example 3: Special Segments in a Circle 11:24
  • Extra Example 4: Special Segments in a Circle 18:09

Transcription: Special Segments in a Circle

Welcome back to

For the next lesson, we are going to go over special segments within a circle.0002

The first type of segments that we are going to talk about is the chord.0008

If you have two chords in a circle (this is AC, and this is chord DE), if you are trying to find the measure of the chords--0015

now, we already went over how to find angle measures from the chords; if these two chords intersect,0026

then we have interior angles; so then, we found that, to find the interior angle, we would have to take the intercepted arc0034

on one side, add it to the intercepted arc on the other side, and then divide it by 2;0046

this is a little bit different; we are not looking for angle measures here--we are looking for side measures, segment measures.0051

If we have two chords, and we want to find the measures of any segments, it is going to be AB times BC, equals DB times BE.0059

You are just taking the product of each of the parts of the chord and making it equal to the product of the parts of the other chord.0078

Again, AB times BC equals DB times BE.0087

The next part: for secant segments, we know that secants are lines that intersect a circle at two points.0097

If we talk about secant segments, then they are just segments that end at the one point, and then it has two endpoints.0107

So, it is still intersecting the circle at two points; but it has endpoints.0119

If we have two secants that are drawn to a circle from an exterior point, and that point here is A,0126

then the product of the measures of one secant segment and its external secant segment is equal0134

to the product of the measures of the other secant segment and its external secant segment.0142

This is a secant segment; this is another secant segment; you are going to take the whole secant segment, AC,0149

times the part of that secant segment that is on the outside of the circle; that is AB.0159

So again, starting over: AC, the whole secant segment, times just the outside part, AB, is equal to this whole secant segment, AE, times the external part, AD.0167

So, it is AC, the whole thing, times the outside part, AB, equals the whole thing, AE, times the outside part, AD.0188

So, it is different than the chord segments, where it is just the part times the part, equaling the part times the other part.0197

The secant segment is the whole thing times the outside part, equal to the whole thing here times this outside part.0204

Let's say that AC is 6, so the whole thing is 6; then, if the whole thing is 6, times AB is 4, and let's say AE is 8,0214

and AD is 3; then when we multiply them together, we know that they have to equal each other: 24 = 24.0238

OK, and then, the next part: this is almost the same as the two secant segments.0252

In this case, we have a tangent and a secant segment.0260

Now, when we have a tangent and a secant segment, it is going to be the tangent, squared,0263

equaling the whole secant segment, AD, times the outside part, AC.0270

Now, from before, the secant segment was the whole times the outside part;0276

well, in the same way, for the tangent, the whole segment and the outside part are the exact same thing.0282

It is like saying the whole AB, times the outside part, AB, equals AD times AC.0292

The whole segment here is AB, times the outside part (is still AB), equal to AD times AC.0301

That is why it became AB2; this is not a new formula--this is the exact same thing as the two secant segments.0316

But it just became AB2, because the whole segment and the outside segment part are exactly the same, so it is just that segment squared.0329

So again, the whole segment, AB, times the outside segment, AB, equals AD times AC.0342

Let's actually try some problems: Find the value of x.0356

These are chords, where they form interior angles; in this case, it would just be the part times the part, equals the part times the part.0363

It is 5 times x, equal to 3 times 7; so this will be 5x = 21; if you divide the 5, then x is going to be 21/5.0374

The same thing happens here: you have chord segments, so it is going to be this part, times this part;0401

just look up here; so it is x + 5, times 6, is equal to x times 12.0408

Don't forget to distribute the 6; so it would be 6x + 30 = x times 12, is 12x.0420

If I subtract the 6x here, 30 = 6x; divide the 6; so then, x becomes 5.0430

Don't forget: if they are chords, then it is just the part times the part, equaling the part times the part.0445

Now, keep in mind: you have seen this in a few different lessons, but if they are asking for the segments, this is the theorem that you use.0450

If they are asking for interior angles, then you are going to have to use the intercepted arc,0460

so it is different formulas, different theorems, for whatever they are looking for.0465

Then, angles...this is another theorem; and then, here we are talking about segments.0471

The next example: Find the value of x; here we have the secant segments at an external point.0480

Then, remember: you are going to take the whole thing, and multiply it by the outside part; set it equal to this whole thing, times the outside part.0490

Now, I am only given these parts; be careful that you don't do the part times the part, like the chords.0504

With this one, you are going to have to do the whole thing, times the outside part.0511

And also, be careful when you look for the whole thing; be careful not to multiply these two numbers together.0516

This is 4x, and this is 5; if I give you a segment, and I tell you that this is 4 and this is 3, what is the measure of the whole thing?0525

It is the 7, because you add them together; you don't multiply them; if this is 4 inches long, and this is 3 inches long, together they are 7 inches long.0537

So, when you are finding the whole thing, don't multiply these numbers; this is 4x + 5; the whole thing is 4x + 5,0546

times the outside part, 5, is equal to this whole secant segment, which is 6x + 4, times the outside part, 4.0557

Then, we are going to distribute this; this is 20x + 25 = 24x + 16.0573

If I subtract this 20x, we get 25 = 4x + 16; subtract the 16; I get 9 = 4x, so then x is going to give me 9/4.0587

Again, do not multiply these numbers together; that is the most common mistake: 4x + 5 is the measure of the whole segment.0610

The next one: here, we have a tangent instead of a secant, but it is the same thing--don't get it confused.0621

I know that there is a separate theorem for it, but just think of it as the exact same thing: the whole segment times the outside segment.0629

So then, here, this is the whole segment; it is not 80; remember: it is the 8, plus the 10, so then the whole segment here is going to be 18;0638

multiply it by the outside segment; it is 8; then, the whole segment, x...multiply it by the outside segment, x.0649

This is going to be 144; that is equal to x2; then, I am going to take the square root of that, so x is going to be 12.0664

Find the value of x: For this one right here, we didn't go over two tangents for this section.0687

But I had this problem here, because I wanted to show you that it is the same concept as the two tangents that we went over in the previous lesson.0699

It applies the same as this section here, because, if you have two tangents that meet at an external point,0709

we know that these two tangents are congruent; so then, for this, x is just equal to 5.0718

That is the answer; but I also wanted to apply the same theorem from this section to this problem.0726

So then, the outside, times the whole thing, is x2 (the whole thing is x; the outside is x);0734

that is equal to...the whole thing is 5, times the outside part is 5; so then, x2 is equal to 25; x is going to equal 5.0742

Although we know that two tangents are congruent if they meet at an external point (from the same circle),0759

all of those secants and tangent...secant, secant, tangent, secant, tangent, tangent--it is the same concept.0769

And then, for this one here, again, the same thing happens: it is going to be the whole thing, which is x + 5,0781, again, don't multiply these is x + 5, times...the outside part is x;0794

it is equal to...the whole thing is 6; the outside part is 6; so 6 times 6, which is 36.0804

Distribute this: x2 + 5x = 36; OK, now, here, you have a quadratic equation, because you have this x2.0812

So, to solve for x, we are going to have to factor this out.0828

I am going to subtract 36, so I can make it equal to 0; now, hopefully, you remember how to factor these.0836

If not, then you can use what I call the x method: you are going to put this number up here,0846

and then this number times this number in front--it is the constant times the leading coefficient;0859

this is 1, so it is just -36; and then, you are going to find two numbers that add to get this and multiply to get this.0864

They add to get that, and multiply to get that.0877

The easiest way to do this is to find a factor pair; we know that, since this is a negative--they are two numbers0883

that multiply to get a negative--that means that one of them is going to be a negative number.0892

Some factor pairs of 36 are going to be 6 times 6, 9 times 4, 12 times 3; which pair is going to give us a positive 5?0898

Now, again: one is going to be a negative number, because they have to multiply to get -36.0917

That means that one of them has to be a negative number; and I know that, if I have a 9 and 4,0923

+9 and -4 are going to add to get 5, and they are going to multiply to get -36.0931

So then, I am going to use these two numbers; I don't have a leading coefficient (or it is just 1),0941

so I don't have to worry about anything else; these are just my numbers.0950

So, it is going to be (x + 9)(x - 4) = 0; and if you want to just double-check, you can FOIL it out.0955

x2...and then this is -4x + 9x; that is +5x; +9 times -4 is -36.0969

Then, from here, now that we have factored, we are going to have to solve for x.0982

We make each of these equal to 0; x + 9 = 0; x - 4 = 0; so here, x is -9; x = +4.0984

Now, we have two answers, but here is the thing: sometimes we can just say that both of those are going to be our answer;0998

but this x is a measure--that is the length of this side right here, this segment.1008

We can't have a negative number; I can't have a pen that is -5 inches long.1017

If you have a negative measure, you would have to cancel that out; you have to cross that out; you can't have a negative measure.1024

If it is finding the distance, finding the length of something, then it can't be a negative number.1031

So then, my answer is going to be just 4.1036

To review for this problem (because we did spend a little bit of time on this): you are going to do the whole thing, x + 5, times x, equals 36.1044

Then, you distribute the x, and then make it equal to 0, because this is now a quadratic equation that you have to factor.1060

Find two numbers that it is going to factor out into; we get (x + 9)(x - 4) = 0.1071

When you solve out for x, then you get -9, and you get +4; this can't be a negative number; therefore, the answer is that x is 4.1077

OK, and then, the final example: we are going to find the values of x and y.1090

I have two tangent circles, two circles that are tangent; they meet at one point.1097

That means that this line is a tangent, also; here is a secant segment and a secant segment.1104

Now, if I look at this circle, this secant, and this tangent, I know that I can use that theorem that says1110

that the whole thing times the outside equals the whole thing times the outside.1122

But I have two variables: x and y are both from the segments of this circle.1127

So, let's look at the other circle: here, I have a secant, and I have a tangent.1136

Now, the whole thing, times the outside, equals the whole thing, times the outside.1142

So, since I only have one variable, I can go ahead and solve using these segments.1148

I am going to solve for y first; that means the whole thing is 18 (10 + 8 is 18), times the outside (is 10), is equal to y2,1153

because that is the whole segment; it is y, and then the outside segment is y; so that is y2.1177

This is 180; that is equal to y2, so y, then, is going to be √180.1182

Now, I can simplify that out; but I want to leave it for now, because I am just thinking of...1197

I know that I am going to have to use this value to find x, and I have a feeling that I am going to have to square this back anyway.1206

So, just leave it like that for now, and then you can simplify it in a second.1214

Here, to find x, it is going to be the whole thing times the outside; that is y2; we know that y is √180, times √180.1219

That is equal to 9 + x (that is the whole thing), times 9.1238

So, here we are going to get...this is 180, equals 81 + 9x; subtract the 81; I am going to get 99 = 9x.1247

Then, divide the 9; x is going to be 11.1273

Now, back to y: we can either simplify that, or we can just use our calculator and find the decimal for it,1286

because we know that 180 is not a perfect square.1298

How do we simplify √180? The easiest way to factor out a radical is to do a factor tree.1302

Let's see, we know that 18 times 10 is 180; this is 9 times 2; this is 5 times 2.1312

When you have a prime number, circle it: here we have 3 times 3, so then all of our numbers are circled.1324

Then, whenever you have a pair, whenever you have two of the same number, you are going to write that on the outside.1333

That is going to come outside; so that is going to be 3...and then we have, also, a 2;1340

so, you are going to multiply those two numbers together; and then, what is not crossed out? It is the 5.1349

That means that that has to stay inside, because the only time it can come out of this radical is when it has two of the same numbers.1354

If it is 32, then that 3 will come out; this is 22, so a 2 came out; so this will be 6√5 is y.1364

Or again, you can just use your calculator and just find the decimal of that.1378

There is y, and there is x; and that is it for this lesson.1383

Thank you for watching