Find the Radius of the Small Green Circle | Step-by-Step Explanation

[@timmysze3363]

A great explanation of this question. Hope you make more videos about algebra.

[@PreMath]

Thank you Timmy dear, I will Glad to hear that! You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@jorgechavesfilho]

Nice. Another way: Construct a right triangle with hypotenuse PF, measuring (1+r). This triangle will be isoceles, with sides (1-r). By Pythagoras, we find the value of the small radius r.

[@theQuantumPenguin]

I did not see any place in the problem description that the 2 lines are perpendicular. The thumbnail is missing a right-angle indicator at the intersection of the 2 lines. ABFC is just declared a square, with no proof, am I missing something here ?

[@jordanh9520]

You're absolutely correct, there is no angle described even implicitly between the two tangents.

[@Morbius907]

This is exactly what I noticed. If ABFC is not a square then the radius of P can be anything between 0 and 1 units.

[@markusbrogle7603]

I love your Videos, but there is a mistake. In the given figure it is not given that the angle of the two Tangents is 90 degree. Of course it looks like that but it is not sure.

[@ciberiada01]

Nice problem and solution, Premath, but the one thing missing was that ∡BAC = 90°. If not, the solution would look more like: r = R·(1/sin(θ/2) - 1)/(1/sin(θ/2) + 1), where R is the big circle's radius and θ is the angle between AC and BC.

[@patrickchapman8583]

I'm confused. Where did we get that the figure ABFC is a Square? Sure it /looks/ like a square. Sure angle ACF and angle ABF are right angles, but I didn't see anything that says the angle CAB is a right angle, or that lines AB and AC are perpendicular, or that the circles are inscribed in a square, etc. What am I missing?

[@mikemason6520]

This problem was solved by assuming the “two lines tangent to both circles” are perpendicular to each other as well. I agree that Patrick’s is correct in that it is not given that these lines are, in fact, perpendicular. Thus the problem can only be truly solved as an equation with the missing angle as a variable, as t is presented.

[@dakkonfury]

THANK YOU! I was thinking the exact same thing. The tangent lines, when connected to the radii of the large circle, form a kite. Now, that kite *can* be a square, but it doesn't have to be, it just has to have two opposite angles of 90 degrees.

[@alexandergutfeldt1144]

Thank you Patrick! You put into words, what bothered me about the solution presented in the video!

[@aakashkarajgikar3935]

3:25, I don't think you need to use the Pythagorean Theorem, you could also look at your 45 45 90 right triangle and notice that the hypotenuse is "√2"!

[@paulpeterson4216]

If the guy has to refer to two theorems to prove that the square he constructs at the start of the video has four equal sides, then he's sure as hell going to go all the way through the Pythagorean Theorem. This is not the guy to listen to if you are capable of skipping steps. He would kill me as a math teacher for "not showing your work" if I did something "wild" like assuming without proof that circles are round or some such.

[@aakashkarajgikar3935]

@@paulpeterson4216 Yes... but I am just giving him less work. I am suggesting less work.

[@robertcaldwell2994]

Yes, this was what I expected him to jump to. 1/2 of root 2 minus 1.

[@robertcaldwell2994]

I missed the corner. Oops.

[@murdock5537]

Awesome! Many thanks, Sir! AH = √2 - 1 = r + r√2 = r(√2 + 1) → r = (√2 - 1)/(√2 + 1) = 3 - 2√2 🙂

[@theoyanto]

Excellent example, I made an error in observation, but now understand my mistake, so I'm happy I learned much from this very instructive example... But you do seem to like your square roots 🤓

[@fongalex6639]

Very clear and without jumping any steps

[@PreMath]

So nice you Fong dear Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃

[@GerardWassink]

I think it can be much simpler. Since the sides of the big square are 1, the diagonal AF is 1,1412 (square root of 2). When we subtract the length of HF (which is 1), and divide it by 2, we get the final answer: aprox. 0,7071. Please tell me why this is different from your answer?

[@GerardWassink]

@@pa28cfi You are right sir, I stand corrected!

[@GerardWassink]

@@pa28cfi LOL…

[@micahwatson9017]

You cannot assume the angle bac is 90 degrees. That’s why Premath’s solution is also incorrect.

[@SureshKumar-nk2ok]

thank u so much sir.i calculated as 2-sq of 2 bcoz is excluded the portion of corner to small circle

[@johnbrennan3372]

I used different construction. Thro P draw a line perp. to BF. Let it meet BF at pt K. That gives rt angled triangle PKF in which (1-r)^2+ (1-r)^2 = (1+r)^2. That gives r= 6+ or - (4 square root 2). Since r must be less than 1, r= 3- (2 sqroot 2).

[@PreMath]

Good job john

[@drgerardfinette4404]

@@PreMath same same

[@TheJamesRedwood]

Like the video itself, you are assuming ABFC is a square, when it could be a kite.

[@johnbrennan3372]

It is easy to prove ABFC is a square.All the angles are rt angles. |AB|=|AC| and |AC|=|BF|.Also |AC|=||CF|. So all sides are equal and all angles are right angles.

[@TheJamesRedwood]

@@johnbrennan3372 What is the proof that all four angles are right angles? |AC| is assumed to equal |CF|, there is no statement in the original diagram that this is so. BAC could literally be any other acute angle depending on the size the circle P.

[@aakashkarajgikar3935]

I really liked this video! Thanks a lot for going over this!

[@PreMath]

You're so welcome Aakash dear! You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@aakashkarajgikar3935]

@@PreMath Thanks! All the best to you too! I also want to tell you that I went to Khan Academy, and I am learning about operations and cool stuff with complex numbers, I can post a link. www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:complex-mul/e/graphically-multiply-complex-numbers?modal=1 If you are interested in checking this out, this is what I am learning on Khan Academy. I am going to take Precalculus math class this coming year.

[@adogonasidecar1262]

How do you prove your starting graph that both circles can actually have the same tangents?

[@hcgreier6037]

Consider an arbitrary radius R for the big circle. Call the radius of the small circle r. Then one gets AF = R√2 AF = r√2 + r + R Thus we find R√2 − R = r√2 + r R·(√2 − 1) = r·(√2 + 1) and r = R·(√2 − 1)/(√2 + 1) Rationalizing the denominator then gives r = (3 − 2√2)·R which is approximately r ≈ 0.17157·R

[@zoranocokoljic8927]

Digonal of every square is a*SQRT(2). You should knowthis by heart. No need to invoke Phytagora.

[@murdock5537]

Indeed!

[@CaesarIanAbatol]

Wish that you posted videos back on 2011, my life in college would be easier. Can you make videos about statistics and advance engineering mathematics? I have a hard time understanding those branches until now.

[@PreMath]

So nice you Don dear for your good words. I've already uploaded quite a few vids on statistics. You may search in my premath channel. I'm planning on making more vids on engineering pretty soon as well. I'm a full time professional and running premath channel by myself... Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃

[@gajavelliprincesses]

Very easy! I did it in no time.

[@PreMath]

Excellent! Thanks Raman dear for the feedback. You are awesome 👍 Keep smiling😊

[@charlesbromberick4247]

Nice job. As an alternate solution, I see a right triangle with sides (1-r), (1-r) and (1+r) which should yield a relatively easy quadratic equation.

[@PreMath]

Great tip! Thanks Charles for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@charlesbromberick4247]

@@vashon100 interestin analysis, mr. v

[@touhami3472]

@@vashon100 there is only one solution : 1-r is a lenght ===> 0

[@touhami3472]

@@vashon100 Charles say : ' right triangle with SIDES 1-r, 1-r and 1+r' So in THIS SITUATION, the unique senario is ' 1-r is a LENGTH ' ==> 0

[@casusincorrabilis1584]

I solved it with the geometric series (inverted). Just need to be careful with chosing the sum, as it is 1/2 times square root of 2 minus 1 and not the whole. As the radius is added twice every time instead of just once.

[@paulk314]

That's how I did it too :)

[@bolder99]

Why is ABFC a square?

[@paulnielander4832]

I agree. There's nothing in the setup that specifies this.

[@ZachGrady]

@premath I have found myself watching these videos, and they are very good, however I have one complaint about this one, and that is you assumed the angle of CAB is 90 degrees. Maybe it was explicitly stated and I missed that.

[@Phill0old]

Since the other 3 angles are 90 degrees what else could it be? Also it is a square.

[@alexandergutfeldt1144]

@@Phill0old But how do you know that the angle CFB=90 degrees? There is no 'proof' of that assumption in the original problem presented, as far as I can see! ACF and ABF are 90 degrees, since the lines AC and AB are tangents to the circle centered on F. ABFC might be 'only' a kite, not a square!

[@herblevinson7133]

Once it was known that the bisector length of large triangle was the aquare root of 2 then equal to 1.414 then 1.414-1=.414/2=value of radius of smaller circle. though the proffessor did do a good job, it could have been shortened.

[@professorsogol5824]

How do we know angle AFB is a 90 degree angle? Can it not be any angle greater than zero and less than 180 degrees?

[@denismilic1878]

The simplest way is for solving this to determine the ratio of radius and diameter of circle + extension to angle point. 1: sqrt(2)+1, length of the smaller circle to angle is sqrt(2)-1 -> r= (sqrt(2)-1)/(sqrt(2)+1) -> 3 - 2sqrt(2).

[@gloubiboulgazeblob]

And for any given R : r = abs(R ( 3 - 2 sqrt(2) )) There's an infinite number of circles, smaller a smaller, recurrence is r_n = r_n-1 * (3 - 2sqrt(2)). In the present case, R = r_n-1 = 1

[@susennath6035]

Excellent

[@gemeni0]

3-2√2 Если угол вверху справа прямой конечно.

[@johnbrennan3372]

Correction to my previous remark : 2r= 6 +or - 4 times sq root of 2

[@PreMath]

Thanks John for the feedback. You are awesome 👍 Keep smiling😊

[@touhami3472]

There is an alternative solution : 1. Draw tangent MN passing through H 2. Consuder the triangle AMN : Area of AMN= AH×MN/2 (1) = sum areas of triangles APM, APN, MNP = r[ AM + AN + MN]/2 (2) Lenghts: AH= AF - FH= sqrt(2) - 1 AM=AN= AH×sqrt(2) MN= 2×AH because AMN is isoceles right on A. Therefore, (1) = r and (2)=3-2×sqrt(2).

[@PreMath]

Good job Themi dear

[@TheJamesRedwood]

Like the video itself, you are assuming ABFC is a square, when it could be a kite.

[@sampathkodi6052]

Here circles are touching each other we can use distance between centers is equal to sum of the radius of the two circles.

[@PreMath]

Thanks Kodi dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@nilton61]

Theresia NO indication that angle A is 90 degrees.

[@SuperYoonHo]

superb

[@canadiannuclearman]

Instead of a set of 2d circles you have 2 sphears. One sphear where its radius is 1 tangent to 3 planes that are perpendicular to one another like the corner of a room. What is the max size spear that will fit in the space between the sphear of R=1 and the 3 planes. Interestingly the radius of the smaller sphear is. r= (sqrt3 -1)/(sqrt3+1) It's the same as your 2d problem but only the 2's becomes 3's. Wow what an interesting connection.

[@philipkudrna5643]

Nice problem covering everything from radius of a circle over the Diagonale in a square up to how to rationalize a fraction!

[@PreMath]

So nice you Philip dear Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@feliksplotnikov6408]

Very simple problem.

[@komodosp]

How about this alternative solution? Extend the AF line so it passes point Z where it intersects the outside of the big circle. |AZ| = sqrt(2) + 1. |AH| = sqrt(2) - 1. Now consider a point Y where AZ intersects the inside of the small circle. The proportion of |AZ| to |AH| is the same as |AH| to |AY|. So (Sqrt(2) + 1) / (Sqrt(2) - 1) = (Sqrt(2) - 1) / |AY|. From that we can get |AY|, subtract it from Sqrt(2) - 1, to get the diameter which is just a short walk from the radius

[@dicboxdicbox6969]

Wow ur Chanel is growing

[@brentsoper345]

So my method used pythag with the 1,1,square root2 triangle to find AH = square root 2-1, then as each circle/ square figures are similar then I solved r/AH = 1/(AH + 2) and this results in r= AH/(AH+2) which gives r = (Square root 2- 1)/(square root2+1)

[@mokshithreddy1634]

thank you sir

[@PreMath]

Most welcome Mokshith dear You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@xtenkfarpl]

It's a bit of a nitpick, but you didn't explicitly specify in the problem statement that the top 'horizontal' line and the left hand 'vertical' line are actually at right angles! How about we generalize to a case where the angle is NOT 90 degrees? Have to crank up a bit of trig then, I think?

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Keep smiling😊

[@theophonchana5025]

Right triangle ABF

[@PanglossDr]

Approximately 1/6 calculated in 15 seconds. It is exactly (root 2) -1 / (root 2 + 1)

[@bluewaterhorizon]

Ho do you come to a conclusion that the angle at CFB is 90 degrees?

[@magicjim1]

@1:10. Yjdidnt explain how you know Segment AF contains P.

[@billtruttschel]

Wertex?

[@potnuruleelarani9335]

You did well however I didn't get the point at PH.from the smaller triangle you got AP=r√2 by Pythagoras theorem then can't we take AH=2(r√2) rather than taking PH=r as we know the diameter in a circle = 2×radius of circle

[@lokesh9228]

1000th like

[@jyotidevi2290]

Thanks

[@PreMath]

Welcome Jyoti dear You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@jyotidevi2290]

Offcourse

[@PreMath]

@@jyotidevi2290 You are the best Jyoti dear. Love and prayers from the USA! 😃

[@cb4920]

So the square of 2 is a constant value to a square like pi is to a circle so we know that the value of f into the corner is the square root of 2. We know after we subtract the radius of the big circle we would be left with .414213562…. as the hypotenuse of a square with sides of .292893219. We also know that .414313562….. is the % by witch the hypotenuse is larger than the side of any square. So multiple the side .292893219 X .414213562…= .121320344 and then because the side of any square multiplied by the square root of two = it’s hypotenuse then .121320344 X 1.414213562… = .171572876 or there abouts, tough to do on my iPad. You see your just figuring the distance from the little circle into the corner and it’s always going to be .414213562% of the hypotenuse.But that’s the way I’d figure it using my math. It would work with a big circle of any radius. I love 1.414213562…. and it’s parts and pieces, I use it like a mechanic uses a crescent wrench. The decimal side of it comes in handy all the time.

[@mikedavis7636]

I find it amusing that the Indian dude calls it the Pythagorean theorem and white dudes are calling it brahamaguptas equation.

[@patrickjacquiot9073]

AH=sqr(2) - 1; power of point A vs small circle : AE²=r²=AH.(AH-2r); that means : r²+2r.AH-AH²=0; positive root = 0,1716

[@abdirazksomo88]

Thnks alot

[@PreMath]

Most welcome Somo dear Thanks dear for the feedback. You are awesome 👍 Keep smiling😊

[@mshanmukhavalli4567]

Interesting sum

[@PreMath]

Thank you Shan dear You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@ddtalks2821]

(2:18) - Square? : This is an assumption. Angles CAB and BFC do not have to be right angles, the sum does have to be 180 degrees. Angle BFC must be < 180 degrees or the tangent lines are parallel (or the tangent lines meet on the other side of the F circle). Quadrilateral ABCF can be a rhombus (having the sides still be equal) or a non square/rhombus. Lengths AC and AB can be longer BFC > 90 degrees) or shorter (BFC < 90 degrees) than 1.

[@baliramkumarsinghpatel1816]

This question asked in INDIAN Railway Exam 2021 Given R = 5 cm r = ?

[@theophonchana5025]

Square ADPE

[@fredsmit3481]

Isn't the diameter of little circle just (square root of 2) - 1 ? If so why not take that and divide by 2 to get radius?

[@TheJamesRedwood]

It isn't. You are forgetting the small gap between the apex A and the small circle also needs to be calculated and then subtracted from sqrt2.

[@AmirgabYT2185]

r=3-2√2≈0,17

[@mustafizrahman2822]

It seems hard to me. I have failed to solve it. But thanks for your solution.

[@PreMath]

Welcome Rahman dear 👍 You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@Areejfz]

I wish you good luck rahman

[@mustafizrahman2822]

@@Areejfz Thanks for your good comment.

[@jtuttle11]

I would probably just split the 90 degree angle in half and measure the diameter of the green circle and divide it in to half.

[@TheJamesRedwood]

There is nothing in the diagram that proves ABFC is a square, it could be any quadrilateral with one degree of symmetry; i.e. a kite with two opposite right angles.

[@johnbrennan3372]

The tangents meet the given diameters of the circle at rt angles So the angles at C and B are rt angles.By joiningB to C you get two congruent triangles ABF and ACF..This gives angles BAC and BFC equal. So each is a rt angle. Therefore figure ABFC is a rectangle.So the opposite sides are equal and all the angles are rt angles To prove this rectangle a square all we need to show is that a pair of adjacent sides are equal in length.if you draw a pair of tangents to a circle from a point outside the circle,the tangents are equal in length.So|AB|=|AC|. So all the sides of figure ABFC are equal in length. Therefore ABFC is a square.

[@TheJamesRedwood]

@@johnbrennan3372 It does NOT give you BAC and BFC being equal, it does NOT prove that BACF is a rectangle, you are assuming that. BACF could easily be a kite, it is not defined in the problem, you have assumed it is not. There is no proof that |AB| = |CF|. A kite would mean that |AB|=|AC|, and |AC| = |BF| as well, this would be true if angle A was 100 and D was 80, or vice versa, or any other combo that equalled 180. In this problem it LOOKS like a square but you are assuming that. Without it being defined at the start, your solution, and pre-math's, could be wrong. Try finding the solution with angle D being 92 degrees and angle A being 88 degrees, you will of course get a different (larger) answer for the radius of the small circle.

[@johnbrennan3372]

What I said was that triangle BAC and BFC are congruent. At the start of the video the tutor says that AB and AC are tangents to both circles

[@TheJamesRedwood]

@@johnbrennan3372 The fact AB and AC are tangent to both circles does not prove that both of those triangles are congruent. Imagine the same diagram with the smaller circle obviously larger than in this one - but still not as large as the circle with centre F. You can draw two lines that are tangent to any two touching circles. Unless the circles are the same diameter, those two lines will meet at a point A. There is only one diameter that will cause the angle at point A to be 90 degrees. It could literally be any other angle less than 180 degrees, depending on the diameter of the smaller circle. Or to put it another way, the diameter of the small circle could be any value between zero and the diameter of the larger circle, depending on the angle at point A. The diagram does nor specify the angle at point A, it just looks like its 90 degrees. It doesn't matter what the tutor says during the video, the information required to solve the problem must be in the diagram, and it is not. Have you had any tutors come up and give you extra required information when you are doing an exam?

[@TheJamesRedwood]

@@johnbrennan3372 Look further into the comments - you will see I am not the first to see this fault in the diagram.

[@theophonchana5025]

Square ABFC

[@Gurpreet_singh_008]

Can't we subtract 1 from √2 to get AH then divide it by 2 to get the radii of green circle 🤔🤔🤔

[@markusbrogle7603]

No this is too long

[@Gurpreet_singh_008]

@@markusbrogle7603 how?

[@markusbrogle7603]

@@Gurpreet_singh_008 AH is not the diameter of the green circle - there is a little gap between A and the circle. Or think: AP is longer than PH

[@Gurpreet_singh_008]

@@markusbrogle7603 oo*

[@duggydugg3937]

neat

[@PreMath]

Thanks dear

[@theophonchana5025]

Radius = 3 - (2 × square root of 2)

[@theophonchana5025]

AP = r × square root of 2

[@stijnnaets6873]

Square root of 2 times x minus x minus 1/2 square root of the square root of 2 times x minus x

[@theophonchana5025]

#Radius

[@VolkGreg]

2(1-r)² = (1+r)² r² - 6r + 1 = 0 r = 3 ± √(3² - 1) = 3 - 2√2 = 0.1716 By inspection, the positive root is too large.

[@theophonchana5025]

#RightTriangle

[@theophonchana5025]

#Pythagoras #PythagoreanTheorem

[@justg4898]

Before I watshing the video: (√2 - 1)/2 That's too easy!!! ...step 4... Wait!, I was wrong! ...nooo I see I must watch the full video

[@theophonchana5025]

c = Square root of 2

[@theophonchana5025]

1; 1; square root of 2

[@TriPham-xd9wk]

(Squaroot of (2)-1)/2

[@magicjim1]

@2:24 You didn't justify that it is a square. 2 right angles doesn't make it a square.

[@blobfish1112]

(1-r)sqrt2 = 1+r.

[@theophonchana5025]

2 × square root of 2

[@theophonchana5025]

AF = Square root of 2

[@nikolaysharapov6298]

1+1=х^2. Х=✓2. ✓2-1. R=(✓2-1)/2.

[@theophonchana5025]

Radius = PD

[@Reddogovereasy]

Instead of teaching math, teach teachers how to teach their students math. If I were going to a university where you taught, I'd sign up very early for your classes.

[@thilagamani3332]

First view

[@PreMath]

Thank you Mani dear You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃

[@theophonchana5025]

Square root of 2

[@theophonchana5025]

Radius = PE

[@theophonchana5025]

r × Square root of 2

[@theophonchana5025]

Radius = FC

[@RPGHouseFabricator]

You lost me with your calculations, but I solved it using a ratio: 2.414 is to 1 as 0.414 is to 0.1715.

[@coachhannah2403]

(Sqrt(2) - 1)/2

[@theophonchana5025]

r = ((Square root of 2) - 1) ÷ ((Square root of 2) + 1)

[@theophonchana5025]

r; r; r × square root of 2

[@theophonchana5025]

Radius = FB

[@alexxx0179]

What the heck was that? Was it English? tuu tata tuut taatat tutu🤣🤣🤣

[@theophonchana5025]

FB = FC = 1

[@wiseguy7224]

OMG Will you ever come to the point ??? You are boring your audience to death...