### Tangents

- Tangent Theorems:
- If a line is a tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency
- In a plane, if a line is perpendicular to a radius of a circle at the endpoint on the circle, then the line is a tangent of the circle
- Know what common external tangents and common internal tangents are using geometric pictures
- Tangent segments: If two segments from the same exterior point are tangent to a circle, then they are congruent
- Circumscribed Polygons: A polygon is circumscribed about a circle if each side of the polygon is tangent to the circle

### Tangents

Determine whether the following statement is true or false.

If ―AB ⊥↔CD, B is on circle A, then ↔CD is the tangent of circle A.

Find a common internal tangent and a common external tangent.

- Internal tangent:↔AB
- External tangent:↔CD

External tangent:↔CD

↔AB and ↔CD are internal tangents, write two pairs of congruent segments.

A pentagon is circumscribed about a cirle if each side of the pentagon is tangent to the circle.

A tangent of a circle is always perpendicular to the radius drawn to the point of tangency.

- ―AC ⊥―BD
- CD = √{AD
^{2}− AC^{2}} = 4 - ―CD ≅ ―BD

CG = 18, EF = 10, find the perimeter of ∆CEG.

- GF = BG, BC = CD, DE = EF
- C = CB + CD + DE + EF + GF + BG
- C = 2CB + 2EF + 2BG
- C = 2CG + 2EF
- C = 2*18 + 2*10

No common internal tangent can be drawn for two concentric circles.

If two circles are concentric, at least one common external tangent can be drawn.

For a circle, the area of its inscribed polygon is _____ smaller than the area of its circumscribed polygon.

*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

### Tangents

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
- Tangent Theorems 0:04
- Tangent Theorem 1
- Tangent Theorem 1 Converse
- Common Tangents 1:34
- Common External Tangent
- Common Internal Tangent
- Tangent Segments 3:08
- Tangent Segments
- Circumscribed Polygons 4:11
- Circumscribed Polygons
- Extra Example 1: Tangents & Circumscribed Polygons 5:50
- Extra Example 2: Tangents & Circumscribed Polygons 8:35
- Extra Example 3: Tangents & Circumscribed Polygons 11:50
- Extra Example 4: Tangents & Circumscribed Polygons 15:43

### Geometry Online Course

### Transcription: Tangents

*Welcome back to Educator.com.*0000

*For this next lesson, we are going to go over tangents.*0001

*Now, remember: tangents are lines that intersect the circle at exactly one point.*0006

*If a line is tangent to a circle, and there is a radius that is also touching that same point, then the radius and this tangent are perpendicular.*0013

*This point that they meet at--this is called the point of tangency.*0030

*If the radius and a tangent meet at that point, then they are perpendicular.*0047

*In a plane, if a line is perpendicular to a radius of a circle with endpoint on the circle, then the line is a tangent of the circle.*0056

*So, it is just the converse of this theorem.*0063

*So again, a tangent, we know, touches the circle at one point; and if the radius is also right there,*0066

*at the point of tangency, then those two are perpendicular.*0075

* And then, the converse is that, if a line is perpendicular to a radius of the circle, then the line is a tangent.*0080

*Common tangents: now again, a tangent has to be intersecting the circle at one point.*0096

*We have the same tangent touching two different circles; it is intersecting two different circles;*0103

*this tangent right here is intersecting this circle at this point, and intersecting this circle at this point; then that is a common tangent,*0111

*because two circles are sharing the same tangent, so it is a common tangent.*0118

*Now, a common tangent can be either external or internal; we know that external means outside, and internal means inside.*0124

*So, when the common tangent (the tangent that the two circles share) is on the outsides of the two circles, then it is an external tangent.*0132

*This is another one; this one is also on the outside, so it is a common external tangent.*0144

*And then, the shared tangents for this one are internal, because it is crossing through between them.*0151

*See how there is nothing here; it is just on the outside, just making a wall with them.*0160

*But here, it is crossing in between them; that is internal, in between--one on that side, and then the other one on the other side of the circle.*0168

*These are common internal tangents; external tangents, and common internal tangents.*0180

*Tangent segments: here is a tangent that is intersecting the circle at this point, point B;*0191

*and this tangent is intersecting at point C; so we have two tangents.*0197

*And those tangents intersect at a point outside the circle, right there; they intersect at point A.*0205

*Then, this tangent segment and this tangent segment are congruent.*0216

*It can't be the whole tangent, because it is going on forever; this segment, from that point of tangency*0227

*to the point where they intersect--that part, that segment right there, is going to be congruent to this segment right here for this tangent.*0235

*Remember that they are congruent.*0249

*Circumscribed polygons: we learned about inscribed polygons; inscribed polygons are when you have a polygon*0254

*that is inside the circle, with all of the vertices, all of the endpoints, touching the circle.*0261

*But this one is on the outside; "circumscribed" means that a polygon is on the outside of the circle,*0267

*so that each side of the polygon is tangent to the circle.*0278

*See, look at this one: this is tangent to this, because it is intersecting the circle at one point.*0283

*This side is tangent to this circle; tangent, tangent, tangent.*0289

*That is circumscribed polygons; I can also say that this circle is inscribed in the pentagon, or the pentagon is circumscribed about the circle.*0296

*Just remember that "inscribed" is inside; so whatever you say is inside--you have to use the word "inscribed."*0311

*The circle is inscribed in the pentagon, and the pentagon is circumscribed about the circle.*0319

*The same thing here: we have a triangle; we have a circle that is inscribed; or I can say that the triangle is circumscribed about the circle.*0330

*And again, the sides of each polygon are tangent to the circle.*0340

*OK, our examples: the first one: Triangle ABC is circumscribed about the circle; if the perimeter of the triangle is 80, find DC.*0354

*You are probably going to get a lot of problems like this, where you are going to have circumscribed polygons; and they all have to do with tangents.*0365

*If you look at this right here, this side of this triangle, it is tangent to this circle, because it is intersecting the circle at that point.*0379

*The same thing happens here: it is intersecting at point E and intersecting at point D.*0389

*All of these sides are tangent to the circle; now, that one theorem that says that,*0395

*if you have two tangents that intersect at an exterior point (we know that this right here and this right here*0403

*are two tangents of the circle that intersect at point B; therefore) these are congruent.*0414

*And then, the same thing happens here: this tangent and this tangent segment intersect at point A; therefore, this part and this part are congruent.*0422

*And then, the same thing happens for those; so for this here, it is like we have three pairs of congruent segments.*0434

*And so, if they give us the perimeter (they tell us that the perimeter is 80), well, if this side is 8, then this side also has to be 8;*0448

*if this side is 12, then this side has to be 12; if this side is x, then this side also has to be x.*0459

*And then, to find the perimeter, we know that we have to just add up all of the sides;*0470

*so then, 8 + 8 is 16, plus 12 + 12 is 24, plus x + x is 2x; that is all going to add up to 80.*0472

*This right here is 40, plus 2x equals 80; 2x...we will subtract 40, and then divide the 2; so x is 20.*0489

*And what are they asking for? DC...well, that is x; so I can say DC is 20.*0504

*OK, again, they want us to find DC; we have a tangent here; AC is a tangent, so this is the point of tangency.*0518

*And the radius is also at the point of tangency right there--the endpoint of that radius.*0534

*Therefore, this radius and this tangent are perpendicular; that was the first theorem that we went over.*0541

*This radius and this tangent are perpendicular, because they are both intersecting at the point of tangency.*0552

*Well, if this is a right angle, then I see here that I have a right triangle.*0560

*They are asking for DC, but I want to first find BC; I am going to find BC first, because I know that BC is a side of the triangle,*0567

*and I can use a triangle to find the unknown side.*0578

*And then, from there, I can look for DC.*0582

*In order to find the missing side of a right triangle, I use the Pythagorean theorem.*0586

*This would be 5 ^{2} + 12^{2} =...let's make that BC^{2}.*0591

*5 squared is 25, plus 144, equals BC squared; this is 169 equals BC squared; therefore, BC is 13.*0602

*Now, if BC, this whole thing, is 13, and I just want to find DC, well, do I know BD?*0622

*If I know BD, then I can just subtract that from 13 and get DC; but how do I find DB?*0633

*Well, BA is a radius with a measure of 5; isn't BD also a radius?*0641

*If BD is a radius, and we know that all radii have the same measure, if this is 5, then BD has to be 5.*0650

*Then, I just subtract 13 from 5, and I get 8; so if this is 5, then this has to be 8; I can say that DC is 8.*0658

*So again, the tangent line and the radius are perpendicular, because they meet at the point of tangency.*0675

*That gives me a right triangle; and then, I use the Pythagorean theorem to find the whole unknown side, BC.*0682

*If they ask me for BC, then that would be the answer; but they are asking for DC.*0692

*So, I found that BD, since that is a radius, is 5, the same as this; and then, subtract it from the whole thing, and I get 8 as DC.*0697

*OK, the next one: Find the value of x.*0712

*Now, for this one, here is a tangent; here is a tangent; and then, here is a segment*0716

*from the center of the circle all the way out to that external point, that point of intersection.*0728

*Then, here is the radius; all I need to look for is x.*0736

*I know that this tangent and this tangent are congruent, because if a tangent and a tangent meet at the same point, then they are congruent.*0745

*So, I can just make this and this equal to each other.*0757

*This segment and this radius mean nothing; they don't mean anything to me; that is all you need.*0759

*Now, you might get a problem similar to this, where they are asking for this segment here.*0765

*Or they give you this segment, and they ask for the tangent segment.*0774

*So, in that case, you can make this radius meet at that point of tangency so that it will be perpendicular.*0778

*And then, this radius will have a measure of 6, and then you can just work with that there.*0789

*But for this problem, we don't need that; here, we just need to make these tangents equal to each other.*0794

*It is going to be 2x - 7 = x + 3; to solve, I can subtract the x, so that would be x; I can add the 7, so x = 10.*0800

*And then, for this one, the radius is 12; they are asking for this whole side of the square.*0821

*We know that it is a square, because all sides are perpendicular, and each side is tangent to the circle, so it has to be a square.*0829

*Now, you can think of this two ways: if you just think of this in a very simple way,*0844

*this side is the same as the diameter of a circle, because it is from one end to the other end, and that is x.*0853

*If the radius is 12, then we know that the whole thing, the diameter, is 24.*0863

*You can also look at this as tangents; and I am explaining this both ways, even though we know that that could be the easiest way to solve,*0868

*because you might have different versions of this kind of problem, where you have a square circumscribed about a circle.*0878

*And just keep in mind that these are all tangent; that is tangent here, and so each of these are going to be congruent to each other.*0887

*Then, remember that tangent segments that meet on an external point are congruent to the same thing here:*0896

*tangent, tangent, congruent, congruent, congruent, congruent.*0901

*Whatever you need to be able to find whatever is that that problem is asking for...*0906

*For our problem here, I can just make x become 24, because if this is 12, then this is 12, and the whole thing is 24.*0914

*If that is x, then this also is going to be x, so x here is 24*0928

*And the fourth example: A regular hexagon is circumscribed about a circle; the radius is 10; find the measure of each side.*0945

*"Regular" means that all of the sides of the hexagon are congruent, and all of the angles are congruent.*0956

*It is equilateral, and it is equiangular.*0968

*Now, if all of the sides are congruent, and we know that they are all tangent to the circle, then, first of all,*0973

*let's draw the radius first, and then we will go on from there: if I am going to draw a radius, I could draw it anywhere.*0988

*They are all going to be 10; but I want to draw it so that it is to the point of tangency, because that helps me out.*0999

*We learned a theorem on that today; so I want to just draw the radius like that--and what do we know about that?*1005

*This radius and this tangent segment are perpendicular; and this is 10.*1016

*Now, I am looking for this whole side; so let's see, that is all we have to work with; all that is given is the radius.*1023

*So, how would we solve this problem? Well, since you know that it is a regular hexagon, you know that each part is broken up*1038

*into congruent parts, into congruent segments or congruent sections of the circle.*1053

*So, if I were to just draw out each radius to each of the points of tangency, then what would each angle measure be?*1060

*I have this right here; I have the radius; the radius has a measure of 10; I don't have anything else.*1088

*So, in that case, I know that this is a right angle; but still, if I want to use the Pythagorean theorem, I still need a second side.*1096

*So, that one chapter on right triangles...if you want to use the Pythagorean theorem, you need two of the three sides.*1105

*If we don't have another side, then we have to have an angle measure, at least.*1117

*So, I can use this circle to find my angle measures, since they are all divided up into equal sections of the circle, because it is a regular hexagon.*1124

*Each angle measure is going to be 360 (because that is the whole thing), divided by 1, 2, 3, 4, 5, 6.*1137

*So, 360 degrees, divided by 6, is going to be 60; that means that this is 60; that means that this is 60 degrees; 60 degrees; 60 degrees; and so on.*1147

*If this is 60, now, that doesn't help me too much; it helps me, but then I need a triangle.*1164

*So, I am going to draw, from this point right here, another segment to the middle, like that; that way, I have a right triangle.*1173

*I'll draw this triangle right here; this is the triangle I am going to work with.*1187

*Now, we found that this whole thing is 60--not that each of these is 60, but the whole thing; this was each of these sections.*1198

*This is 60; then, this has to be 30; now, if I want to draw that triangle out again, just so that it is easier to see,*1206

*it is a right triangle; this is 30 degrees; this is 10; what is this angle measure here? This is 60.*1223

*Now, if you want, you can go ahead and use Soh-cah-toa; and we know that we use Soh-cah-toa when we have angles and sides.*1233

*The Pythagorean theorem we only use when we have sides; that is it--only sides--nothing to do with angles.*1244

*But it is all for right triangles, of course, but only when it comes to sides.*1250

*Soh-cah-toa, we use when we have a combination of angles and sides.*1256

*We can use Soh-cah-toa; now, we are looking for this right here, x, because if we find this,*1261

*then we can just multiply that by 2, and we will find the whole side.*1268

*Or we can use special right triangles; now, special right triangles are either 45-45-90 triangles or 30-60-90 triangles.*1274

*If you have a 30-60-90 triangle, this is n; the side opposite the 60-degree angle is going to be n√3;*1294

*and then, the side opposite the 90 is going to be 2n.*1307

*This, the side opposite 30, is going to be n; the side opposite the 60 is going to be n√3; and this is going to be 2n.*1315

*Now, we want to look for this here; what are we given?*1330

*We are given that the side opposite the 60 is 10, so that means I can make 10 equal to n√3.*1334

*Again, in a 30-60-90 triangle, the side opposite the 30 is n; the side opposite the 60 is n√3; and the side opposite the 90 is 2n.*1347

*This is the rule for special right triangles; now, what is given?--this side right here, the side opposite the 60; they gave us that it is 10.*1358

*So, in that case, you can make 10 equal to n√3, and that will help you; and then, you just solve for n.*1367

*So, here I divide this by √3; n = 10/√3.*1375

*Now, here, we have a radical in the denominator; I don't like to have (you shouldn't have) a radical in the denominator.*1384

*So, you want to go ahead and rationalize it; that means that we want to make it so that that denominator doesn't have the square root.*1395

*I can multiply this by itself, √3/√3; now, I have to multiply the top and the bottom by √3,*1404

*because this √3/√3 becomes just 1; this is 1; √3/√3 = 1.*1415

*So, I have to multiply this by this, so that the radical will go away.*1423

*This will become 10√3/3; so, this side right here, n, is 10√3/3, which is also x; that is this side.*1430

*Now, I want to find the measure of each side; that means that each side is this whole thing right here.*1450

*I can take 10√3/3, and I am going to multiply it by 2 times 2; and then, I am going to get 20√3/3, and that is my answer.*1457

*Now, if you want, you can use your calculator; you could just make that into a decimal...or that should be fine, too.*1479

*Again, just to explain how we did the problem: it is a regular hexagon, so I know that,*1488

*if I draw a radius to each point of tangency, then the circle will be divided up into equal congruent angle measures,*1499

*which makes each of these 60; this whole thing is 60.*1509

*Then, since I want a triangle, because with triangles, I have a lot to work with; I drew a segment from this point all the way to the center.*1514

*And then, since this whole thing is 60, and this is 30, this became a 30-60-90 triangle, which is also a special right triangle.*1529

*The side opposite 30 is n; opposite 60 is n√3; opposite 90 is 2n.*1540

*So then, I want to look for the side opposite the 30; this is given to me, so make that equal to n√3; solve for n.*1546

*That is this side right here; multiply that by 2 to get this whole side, and that is 20√3/2.*1560

*That is it for this lesson; thank you for watching Educator.com.*1572

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