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  • 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


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
Internal tangent:AB
External tangent:CD

AB and CD are internal tangents, write two pairs of congruent segments.
OD ≅ OA , OB ≅ OC .
Determine whether the following statement is true or false.
A pentagon is circumscribed about a cirle if each side of the pentagon is tangent to the circle.
Determine whether the following statement is true or false.
A tangent of a circle is always perpendicular to the radius drawn to the point of tangency.
CD and BD are tangents of circle A, AC = 3, AD = 5, find BD.
  • AC ⊥BD
  • CD = √{AD2 − AC2} = 4
  • CD ≅ BD
BD = CD = 4.

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
C = 56.
Determine whether the following statement is true or false.
No common internal tangent can be drawn for two concentric circles.
Determine whether the following statement is true or false.
If two circles are concentric, at least one common external tangent can be drawn.
Fill in the blank with sometimes, never or always.
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.



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

Transcription: Tangents

Welcome back to

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 tangency0227

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 polygon0254

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 here0403

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 52 + 122 =...let's make that BC2.0591

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 segment0716

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 240928

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 up1038

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