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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

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

BF = 12, CF = 4, EF = 6, find DF.

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

- BF
^{2}= DF*EF - BF
^{2}= 12*6 = 72 - BF = 6√2

BF = 6, EF = 3, find DE.

- BF
^{2}= DF*EF - 6
^{2}= DF*3 - DF = 12
- DE = DF − EF
- DE = 12 − 3 = 8

- BF*CF = EF*DF
- CF = [EF*DF/BF]
- CF = [8*5/10] = 4.
- x − 2 = 4

- DF = 2r + EF
- DF = 2*4 + 6 = 14
- BF
^{2}= DF*EF - BF
^{2}= 14*6 = 84

find x.

- DF = 10 + 4 = 14
- BF = x + 6
- BF*CF = EF*DF
- (x + 6)*6 = 4*14
- x = [4*14/6] − 6

AE = 5, AD = 9, find AB.

- AC
^{2}= AE*AD - AC
^{2}= 5*9 = 45 - 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.

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

### Geometry Online Course

### Transcription: Special Segments in a Circle

*Welcome back to Educator.com.*0000

*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 arc*0034

*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 equal*0134

*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 AB ^{2}; this is not a new formula--this is the exact same thing as the two secant segments.*0316

*But it just became AB ^{2}, 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 x ^{2}; 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 x ^{2} (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, x ^{2} 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

*times...now, again, don't multiply these together...it 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: x ^{2} + 5x = 36; OK, now, here, you have a quadratic equation, because you have this x^{2}.*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 numbers*0883

*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

*x ^{2}...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 says*1110

*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 y ^{2},*1153

*because that is the whole segment; it is y, and then the outside segment is y; so that is y ^{2}.*1177

*This is 180; that is equal to y ^{2}, 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 y ^{2}; 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 3 ^{2}, then that 3 will come out; this is 2^{2}, 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 Educator.com.*1387

0 answers

Post by Jorge Gonzalez on December 7, 2016

Great lesson!! Thank you plenty.