Can You Find the Radius of the Small Yellow Circle? | Step-by-Step Explanation

[@India-jq7pi]

Thank you sir

[@PreMath]

So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@osvaldoperotti9662]

The result is 2/9. Check your calculations

[@briank10101]

@@PreMath You bolloxed it up in the 2nd last step. As my Mommy says check your math. 466 should be 46.6. But then last step magically corrected itself as you transcribed it from the solution handbook. 😳 😳 😔

[@N1CaiquelS2]

@@PreMathU're a excellent teacher, i like ur videos so much, i see you fromBrazil🇧🇷❤️ So i thikn u can use this formule: 1/√r = 1/√4 + 1/√2

[@pritamyadav17]

@@PreMath sir plzz allow to download also..plzzz

[@andrewjames6676]

There's a mistake in the decimal point : (4 + 2sqrt2)e2 = 46.627 Anyway, I enjoy your explanations - I'm 80 and it keeps the grey cells busy!

[@2dboys230]

Damn sir! I am a little inspired by your story

[@pezosdrare3291]

Thank you sir I was wondering what was going wrong

[@UbiytsaEditz]

@@awamy91 immature kid

[@UbiytsaEditz]

@@awamy91 at least he is better and doesnt waste his life in religion

[@birindersingh2215]

Sir, you are big inspiration for all of us..... please ignore few rascals!!

[@PiyushSharmaMotivation]

There is a slight mistake in the points while calculating.. Instead of 466.27 it should be 46.62

[@toyuryuz]

It's not slight,it ia a big mistake

[@BinuJasim]

@@toyuryuz no it's a small mistake. If I were evaluating, i would only reduce 1/2 mark for it like from a 6 mark question. Because all the hard steps are done perfectly.

[@akshatpandey9141]

@@BinuJasim agree.

[@toyuryuz]

@@akshatpandey9141 indians supporting indians

[@mijanahmed4633]

@@toyuryuz yes this is brain breaking mistakes.

[@اممدنحمظ]

تمرين جميل شرح مفصل مرتب . وبارك الله فيكم وعليكم والله يحفظكم ويرعاكم ويحميكم وينصركم جميعا . شكرا جزيلا لكم د تحياتنا لكم من غزة فلسطين .

[@wwoods66]

15:00 You could have taken it a little further before pulling out the calculator: r = 12 - 8√2

[@GO_GO_60]

PreMath is idiot.

[@jarikosonen4079]

Maybe the denominator should been rationalized... looks correct though (except the 466.27 should been written as 46.627?) All of this math videos from all the publishers seem have spelling errors some cases.

[@Maahi_xgod]

15:00 (4+2√2) ² =46.6274

[@edziobydlak3096]

Your method can be used only if the radius of the blue circle is twice as big as the radius of the red circle (the tangent needs to be parallel to the base line). There is a similar method that works for all circles.

[@MarieAnne.]

@@alexandertrang6549 By construction, FM is always equal to AC, since FM and AC are horizontal lines, and FNA and MC are vertical lines.

[@larswilms8275]

The tangent to the circle theorem is not needed. The fact that line FM is tangent is a coincidence and makes it, only slightly, easier. Point F could also have been drawn under point M at the same height as point N. Now a right angle triangle can be constructed. Just like before with the yellow circle FN is the difference between the two circle radia. NM is the sum of the two radia. Pythagoras will give you the length of FM

[@darrylmelander6984]

The fact that the horizontal line in the final triangle is tangent to the red circle is coincidental, doesn't actually make it easier, and in my opinion was a distraction he shouldn't have even mentioned. Getting the length of FN is obvious and trivial.

[@sahil_animefan]

But my answer is 0.25😥 ..why

[@polupridurok]

How did you determine that a tangent to NF hit point M?

[@specialcarecars]

Didi you get the answer?

[@kirin347]

Decimal point is off by 1 on th step before the end, but corrected on final result.

[@j.r.1210]

Yes. (4 + 2*2^.5)^2 = 46.627, not 466.27.

[@PreMath]

So nice of you dear! Thanks for your honest feedback. You are awesome 👍 Take care dear and stay blessed😃

[@waeelrihan3072]

@@j.r.1210 its 46.624

[@g.gaston3302]

@@j.r.1210 But 32 / 466.27 is 0.0686, not 0.686. So, it cancels out the first decimal mistake and his answer is still correct.

[@BSnicks]

@@italixgaming915 There is no need to do all these considerations. After finding sqrt(32) = sqrt(8r) + sqrt(16r), it is easy to easy to get to r = 4/(3+2sqrt(2)) = 12-8sqrt(2) by multiplying with the conjugate denominator.

[@princejag]

My day never completes without watching your teachings!!!!!!!!!!

[@PreMath]

Wow, thank you my dear friend Jagadeesh! Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻

[@mathsislove197]

Fax bruh fax

[@kutyuvadi6858]

(4+2sqr(2))^2=46.627 and not 466.27, but the final result is OK. The explanation is correct, although a little too slow. Thanks.

[@ОляЛюба]

Thank you. Very well.

[@rclrd1]

Why on Earth use a _calculator_ to get an _approximate_ answer? The exact answer is simply r = 4/(3 + 2√2).

[@mathsislove197]

The reason is that we want to get exact and approximate answer

[@kenwen7791]

Is it not also 12 - 8√2 ?

[@l3igl2eaper]

@@kenwen7791 That is the rationalized + simplified answer. More appropriate in a math setting.

[@rayrash1]

In one of your videos, you had established that the distance between their contacts with the common tangent of two osculating circles is d = 2 \sqrt r_1 \times r_2 , where the rs are the radii of the two circles. Applying this directly to this problem thrice leads to your answer. If r is the radius of the small circle, x the distance between the points of contact of the radius 2 circle and the small circle, X that been the points of contact of the small circle and the one with radius 4, we get x = 2 \sqrt (2r), X = 2 \sqrt (4r) and x+X = 2 \sqrt (8). This directly gives r = 0.686.

[@pedroreis6316]

Very nice! But It should have been prooved that F, N and A are collinear and therefore you could say that |FM|=|AC|, otherwise, It can't be assumed just by the image. In fact they are collinear because the height in relation to AC of 2 times the radius of the left circle is exactly the radius of the right circle, then when you connect with a perpendicular line (tangent) that line will be parallel to AC and almost |AC|=|FM|

[@brenotavares4543]

I thought the same thing

[@wadsalman8591]

I think not necessary...line FA Is perpendicular to line AC with length 4 m and points And F are tangency points .And radius MC is 4 m. and perpendicular to line AC . So AFMC is rectangle.

[@prabuddhaghosh2543]

The essence of the problem is exactly the point you make. The tangent at F has to pass through the Center of the larger circle M and it’s slope shall be equal to the slope of the resting plane as highlighted by you. Thank You.

[@fedest]

You are right Pedro, there’s a proof missing there. but rather than making that proof, the generalization of the solution should build that triangle without caring for the tangency to the red circle, and define F as the point where a parallel to AC that pases through M intersects the projection of AN. Then the length of FN would then be calculated as MC-NA = 4-2 = 2, which happens to coincide with the radius of the red circle and that’s why F is the tangency point, but that fact is irrelevant. Actually a more intuitive way to build that triangle would be to take the parallel to AC that passes through N and situate F at its intersection with MC.

[@alf8718]

Which brings us to the basic question - can this method be used for a different radiuses or just for some special cases!

[@alokekumar6326]

wonderfully explained with the help of simple geometry and algebra thank you very much

[@aggsar4411]

At 11:40 i think you missed a point we had to explain (i know it's minor but still good to mention). We know that they are 90 degrees (because tangent and radius of same circle) but the question is why does the tangent cross the centre of the other circle and that is because the diameter of the small circle is equal to the radius of the big circle (might be obvious but always good to mention) and that's how we know FM is parallel to the line below

[@mihailsprincean5289]

Thank you. Good explanation. Not just "you can see on the picture"

[@Esmio06]

Dropped it matter the ratio of the circles? E.g. if a circle with radius of 1000 was on the same place as a circle of radius 2 and touching at only one point, would the center of the large circle always intersect the radius of the small circle on a tangent?

[@p.m.6594]

Solving this problem with law of cosines for triangle NPM is also really convenient, you calculate cosine of angle NPM using trigonometric equations for finding cos(180-a-b), where cos and sin of angles "a" and "b" are easy to calculate using two rectangular triangles of NXP and MYP where points X and Y are below circles' centres and lay in a distance of respectively 2-r and 4-r to points N and M. Then you plug calculated cosine into law of cosines for traingle NPM and end up with equation of one unkown.

[@tylerwright6380]

One of the best geometric problems you have presented. Thank you so much!

[@yigit4044]

this is middle school level😂

[@waeelrihan3072]

The basic idea is great Make the radius whose value you want to know one of the values in one of the sides of a right-angled triangle You are a genius, man

[@bigkamran]

Nice!

[@vinaymalwiya218]

Should be 46.6 in place of 466. Final result is however correct.

[@davidvolland7250]

I went over my my math 10X and agree with you he is off a decimal place

[@alexalbert3026]

square root of 2 is about 1, so 2sqr(2) is about 2 and is definitely not more than 4, so 4 + 2sqr(2) will be definitely not more than 8, it is so clear !!!

[@ItsMagzZz]

I was wondering how 4 + (2 times the initial value of 1.414) ended up near 70 lol. Makes much more sense now!

[@satyapalsingh4429]

If radii are a,b&c (a

[@timothybrown390]

Looking at the question gave me anxiety, weird, right? Watching him work through it was satisfying. So much learned in school, stored away and forgotten about.

[@TheFlax33]

Exactly

[@ПерстФомы]

Let R1 be the radius of the blue circle and R2 the radius of the red circle. First, I proved that for any circles with radii R1 and R2 touching each other (as in the figure), the length of the segment of the common tangent AC for them will always be equal to AC = 2√R1 * ✓R2. Now the same is for blue and yellow touching circles with radii R1 and r is a segment of a common tangent for them BC = 2√R1 * ✓r. Likewise for red and yellow contiguous circles with radii R2 and r is a segment of the common tangent for them AB = 2√R2 * ✓r. Substitute into the equation AB + BC = AC and we get: 2√R1 * ✓R2 = 2√R2 * ✓r + 2√R1 * ✓r Hence we have ✓r = √R1 * ✓R2 / (√R1 + ✓R2) This is identical to that obtained by the author of the video.

[@grinfacelaxu]

Exactly!

[@jaishankarfans9029]

I didn't understand your method *How ??* - for any circles with radii R1 and R2 touching each other (as in the figure), the length of the segment of the common tangent AC for them will always be equal to AC = 2√R1 * ✓R2.

[@hassanmukhtar3326]

Can you explain? How does the tangent at F from the red circle meet the center of the blue circle and be parallel to AC?

[@specialcarecars]

Did you get the answer?

[@adampiechuta5774]

11:10 Why did you create a triangle NFM this way? Only because radius of big circle is 2 times bigger than middle circle's radius causes that angle F is 90, in another cause couldn't be. Angle F should be on the line between point M and point C and you could have the same triangle but be sure that angle F is 90.

[@massimookissed1023]

Yep.

[@larswilms8275]

For the right angle triangles you constructed and calculating AB and BC and FM, it helps to rewrite the pythagorean theorem from a^2 +b^2 = c^2 to a^2 = c^2 - b^2. You now have a difference of squares on one side of the equation. This can be written as a^2 = (c-b)(c+b). If c and b are the known adjacent and hypotenuse , like in this example, then the length of the remaining adjacent side is easily determined. the final answer can be simplified further: (4 + 2sqrt(2))^2 * R = 32 expand the square (16 + 2 * 4 * 2 sqrt(2) + 8) R = 32 divide by 8 ((2+1) + 2 sqrt(2)) R = 4 divide by (3 + 2 sqrt(2)) R = 4 / (3 + 2 sqrt(2)) multiple by 1 as (3 - 2 sqrt(2)) / (3 - 2 sqrt(2)) R = 4 (3 - 2 sqrt(2)) / ((3 + 2 sqrt2)(3 - 2 sqrt(2)) the denominator is now of the form (A+B)(A-B) = A^2 - B^2 (3 + 2 sqrt2)(3 - 2 sqrt(2) = 9 - 2*4 = 1 R = 4 (3 - 2 sqrt(2)) = 12 - 8 sqrt(2) R is approximately 0.686

[@BRINDANI2000]

So awesome, thats the way that I did it too. Old college brain cells waking up.

[@Rzoockxz]

By using the Pythagorean theorem, we can quite easily get a much more general solution R = r1 * r2 * (r1 + r2 - 2sqrt(r1 * r2)) / (r2 - r1)^2 where r1, r2 are the radii of the outer circles. And for the degenerated case when r1 = r2 = r, we have R = r/4.

[@everlast6128]

Nice rewriting of c^2-b^2 as (c+b)(c-b), didn’t even think about that

[@tontonbeber4555]

I've always been null in geometry so I pose equations of distances between centers of the circles : coord centers of 3 circles (0 2) (a x) (b 4) b2+4=36 => b2=32 => b=4V2 a2 + (2-x)2 = (2+x)2 => a2 = 4.2x = 8x (b-a)2 + (4-x)2 = (4+x)2 => (b-a)2 = 8.2x = 16x Thus (b-a)2 = 2a2 => (b-a) = V2 a => 4V2 = (V2+1)a => a = 4V2(V2-1)/1 = 8-4V2 = 4 (2-V2) a2 = 16 (6-4V2) = 8x x= 2(6-4V2) = 12-8V2 = 0.686

[@tehatte]

Thanks for the interesting problem! I found a flaw in your explanation. I don’t think you can assume FM is tangent to the red circle, or if you draw the tangent line to the red circle at point F it will go through point M. It is only true in this apecific problem because it just happens that AF, the red circle diameter, is 4 and is same as CM, the radius of the blue circle, which makes ACMF a rectangle, thus the angle NFM is right angle, thus FM is the tangent line.

[@rustyblade22]

Had the same thought. Ideally should have repeated the process and taken ACMN.

[@mathbum33]

Their radii stipulate the tangent lines, since the diameter of one is the radius of the there it has to be correct

[@OVRxNxOUT]

Sure you can assume FM is the tangent to the red circle as the red circle’s diameter is the radius of the blue circle. If we were given the radius for the blue & yellow circles and had to solve for R of the red circle then you could not assume that FM would be a tangent to red. < I only say this because that was my first thought when he displayed triangle NFM. I thought, “Wait, how does he know FM is the tangent?” I had to pause for a second to then just realize I’m slow sometimes & we were given the radius of both red & blue & the D of one = R of the other. I think because those values get taken off the screen & so many other numbers & calculations happen that one may momentarily forget a small detail like that.

[@tehatte]

@@OVRxNxOUT As I explained in my comments, same as you, is because FA just happens to be equal to MC, thus FM is parallel to AC, thus FM is perpendicular to FA, thus FM is the tangent line. What I meant by “can’t assume” is he didn’t explain that logic. He just drew the line FM and put the symbol for right triangle there without explaining. Some people, including you and I, may question it, and may not find out why.

[@OVRxNxOUT]

@@tehatte good point.

[@georgiosnektarioslilis4863]

Great video. I think what is missing is to prove that FM is tangent. It is implied as the radius of the big circle is double the radius of the left circle. You have to mention that for completeness.

[@TheDGomezzi]

It’s not important that FM is tangent. Even with a much bigger circle, you would simply draw F at a higher spot, and as long as you drew FM parallel to the ground, it still works.

[@DavidNitzscheBell]

@@TheDGomezzi but if you drew F at a higher spot, you wouldn't know the precise length of NF. It only helps us here because we know that NF is a radius of the circle.

[@DavidNitzscheBell]

Exactly. Point M is 4 units from the baseline. Point F is 4 units (diameter or two radii) from the baseline. Therefore, the line segment FM is parallel to the baseline. Therefore, the angle is 90. Therefore it's tangent. I think he got his reasoning backwards.

[@TheDGomezzi]

@@DavidNitzscheBell Nah, you can easily get the length of NF by subtracting NA from MC.

[@billyallen7186]

You can also solve this question with : 1/√biggest radius + 1/√middle radius = 1/√smallest radius

[@tanyaerskine7657]

That was really cool to watch. Thank you!

[@PreMath]

Thanks Tanya for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@davidallen3990]

Crucial to answering the question is to recognize the relationship of the circles touching each other and sitting on a base plane, the rest is all Pythagorean theory. However if one circle wasn’t twice as big as the other it doesn’t seems like that formula would work.

[@straighta9539]

Oops! I love and appreciate your work. The content you put out is outstanding (pace, step-by-step clarity of explanation, use of technology). Given the quantity of videos you produce, there are bound to be some oopses (don't I know it!). So, not to criticize, but rather to correct an oops, you put 466.27 instead of 46.627. Obviously you used the correct value when you divided 32, because the value of the radius did come out correct.

[@spillovereducationtutorial3228]

You have explained the concept of the question very well and I believe the students have understood how to do the question

[@jootpepet]

He showed us the solution, but he never explained HOW he came up with it. I learned nothing.

[@THECUBER2023]

@@jootpepet Totally agree

[@ROCCOANDROXY]

I generalized the problem using R1 and R2 for the radii of the two larger circles and r for the radius of the smaller inner circle. Your problem was a special case where say R2 = 2R1. Using the Pythagorean theorem and the law of cosines I obtained: r = (R1 * R2)/(R1 + R2 + 2 * sqrt(R1 * R2))

[@boguslawszostak1784]

Why complicate what is simple. We have three circles, each pair are circles tangent to each other and to a given line. It is enough to solve it once using the Pythagorean theorem: (ri + rj) ^ 2 = (ri-rj) ^ 2 + distij ^ 2 distij = sqrt ((ri + rj) ^ 2- (ri + rj) ^ 2) = sqrt (4 * ri * rj) = 2sqrt (ri * rj) dit12 = dist13 + dist23 after substitution 2sqrt (r1 * r2) = 2sqrt (r1 * r3) + 2sqrt (r2 * r3) sqrt (r1 * r2) = sqrt (r1 * r3) + sqrt (r2 * r3) = sqrt (r3) * (sqrt (r1) + sqrt (r2)) sqrt (r3) = sqrtr (1 * r2) / (sqrt (r1) + sqrt (r2)) so r3 = r1 * r2 / (sqrt (r1) + sqrt (r2)) ^ 2 for r1 = 2; r2 = 4 r3 = 2 * 4 / (2 + sqrt (2)) ^ 2 = 8 / (6 + 4sqrt (2)) = 12 - 8 sqrt (2)

[@navendulad5093]

The correct solution

[@deenulazarus5454]

Still got the result... Loved the explanation... Couldn't stop but wonder how people in olden days used to calculate by hand ... Measuring the sides, lengths and angles

[@arturgilson489]

By the formula X = (R × r )/ (R+r+2√Rr) Where X is radius of circle yellow and R the radius of circle Blue and r the radius of circle red. X = (4 × 2) /(4+2+2√4×√2) X = 8/(6+4√2) X = 4/(3+2√2) X=4/(3+2√2) × 3-2√2 / 3-2√2 X=(4×( 3 - 2√2))/(3)²-(2√2)² X=(12-8√2)/(9 - 8) therefore X = 12 - 8√2 X ~= 0.68629

[@amarsa24]

thank you. it looks work only big circle radius is two times of medium circle radius. if it doesn't calculation is little different. but idea is same. nice one.

[@lesliedion8289]

It sharpens my algebra and trigonometry understanding.. Thanks for the video

[@murdock5537]

Awesome task, excellent way to solve it, but for some may be a bit confusing (see some comments below; 46,627 instead of 466,27, for example...). Should have explained that FM = AC = AB + BC. A slightly different way: (AC)^2 = 36 - 4 = 32 → AC = 4√2 (AB)^2 = (r + 2)^2 - (r - 2)^2 = 8r → AB = √8r = 2√2r (BC)^2 = (r + 4)^2 - (r - 4)^2 = 16r → BC = 4√r AB + BC = AC = 4√2 = 2√2r + 4√r → √r = (4√2)/(2√2 + 4) → r = 32/8(3 + 2√2) = 4(3 - 2√2) ≈ 0,68629 🙂 btw: tan⁡(θ) = MD/BC = (4 - r)/(4√r) = (4 - 4(3 - 2√2))/(4√r) = 8(√2 - 1)/8(√2 - 1) = 1 → θ = 45° → MD = BC =8(√2 - 1) and: tan(φ) = EN/AB = 2(4√2 - 5)/4(2 - √2) = (1/4)(3√2 - 2) → φ ≈ 29,28°→ θ + φ =74,28° → NPM = 180° - (θ + φ) = 105,72° ≠ 90°

[@MrGelesh]

11:36, How is AFM a right angle tryangle? Or in Other words.. Why is it nessary that, for red circle,tangent from the point F would meet M, the centre of Blue circle

[@royaleminiatures7150]

Great breakdown! I really like how you make your initial sketches refined. Great quality videos thank you!

[@TheFlax33]

Outstanding video. Happy Holidays...I love my PreMath mentor. I am thankful. All the best from the many of us out here who follow you.

[@baxperiencemonastry]

Sir the value of (4+2√2)² is 46.6274 So it will be r=32÷46.6274 r=0.686👍👍 Sir please make videos related to maths questions which are asked in Banking Exams (SBI PO, IBPS PO)🙏🙏

[@dvash5361]

В натуре, у Альцгеймера проблемы с арифметикой!

[@johnnes-777]

@@dvash5361 нет проблем у него - только смекалка!

[@dvash5361]

@@johnnes-777 какая, нафиг, смекалка?! Нам в школе за такую "смеккалку" мигом "пары" ставили без амнистии! Накалякал циферков и написал правильный ответ - чехлись, не проканает! А этот буратино ляпает запятые как бог на душу положит, и всё о,кейно! Некоторые (ну совсем некоторые!) указывают ему на ляпы, он благодарит, хвалит за внимательность и нифига не исправляет. Бамбук.

[@여영철-v4h]

옥의티

[@JeanBaptisteEmanuelZorg]

Can I use a ruler?

[@ghmaxiron2639]

Hi. At 14:42 if squared root of 2 equal 1.414 , then 4+2✓2=6.828 And (4+2✓2)^2=46.62 Thanks for video

[@PreMath]

Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@RMF49]

I did it in my head by the same method. I first generalized that given adjacent circles with radius a and b the horizontal distance between their centers is sqrt((a+b)^2 - (a-b)^2) or sqrt(4ab). Once you have that it’s ez pz.

[@nigelsw55]

Hi, Essentially the same method as me but I left my answer in terms of square roots. I just find it neater unless of course you require the answer to so many decimal points. I enjoy these problems because as an ex maths tutor I would give similar problems to my colleagues.

[@paradoxycycline]

How can we assume that FM is a tangent of the circle having centre N?

[@TheFlax33]

nice......it time for me to revisit trig and geometry. TY for keeping my mind fresh.

[@borelm1]

Step 6 is not needed since FM = AB + BC also it must be noted that for each d = horizontal projection of the distance between centers of 2 circles we have by Pythagorean theorem we have d= 2sqrt(R1R2) where R1 is first circle's radius and R2 second circle's radius. since horizontal projection distance between two large circles is equal to the sum of horizontal projection distances from the small circle to each larger circle we easily obtain Sqrt(R1*r) + Sqrt(R2*r) = Sqrt(R1*R2) where r is the radius of the small circle and R1,R2 are radii of larger circles.

[@tomclemens4761]

You do not need the tangent to a circle theorem at 12:00 if you draw the triangles the same way as you did the first two times! And there is a HUGE simplification you can do with the difference in perfect squares to save days on the algebra.

[@NareshKumar-i3z8g]

Do the tangent of small circle always meet the big circle always?

[@rono1076]

How do you know that F and M are equidistant to the line AC? Was that a given?

[@OldSloGuy]

Simple inspection of the diagram should raise suspicions. There is a misplaced decimal point near the end. 466,27 should be 46.627. The answer is correct however.

[@venantiussheela226]

Sir How can we say FM is tangent to the medium circle?

[@greenlampshade8909]

It is a special case of tangent circles that are also tangent to a common line, where one circle is twice the diameter of the other. A line parallel to the common line through the center of the larger circle, intersecting a line drawn perpendicular to the common line through the center of the smaller circle, will be tangent to the smaller circle at point of intersection. Any other configuration requires bit more effort: kzbin.info/www/bejne/f5TKqpZ9iNZsi9U

[@MehrdadIrani24]

@@greenlampshade8909 That needs to be mentioned, otherwise it is not complete.

[@DDX01]

Very nice videos I am realy enjoying your videos👍👍👍

[@floardeche6972]

32/466 is not equal to 0,6 but the mistake was before :) Very nice demonstration

[@ramanma9915]

At the outset the proof is elegant. But, it is assumed that MF is the tangent to the red circle, whereas, we need to prove it. Join AN and extend it to meet the circle at F. Join FM. Now AF=MC=4 and AF parallel CM. Hence, ACMF is a rectangle and clearly, MF is tangent to the red circle.

[@emmanuelraj...]

Sir, can you please explain how FM is parallel to AC (i.e have same length) ? Is there a theorem that suggests that if 2 circles placed tangent to the same line, then a tangential line from opposite side one circle to the center of the 2nd circle is parallel to the first line ? Will this be true if both the circles are off same radius ?

[@massimookissed1023]

Point F is 2 radii of 2 above the horizontal ground line, Point M is 1 radius of 4 above the horizontal ground line. A more general case would be length NF is radius of blue minus radius of red.

[@bob76451]

This is a very special case where the blue circle is exactly twice as large as the red circle. Because of this, the diameter of the red circle is exactly the length of the radius of the blue circle

[@emmanuelraj...]

@@bob76451 oooh ok. That explains. Thank you for the crisp yet precise explanation

[@emmanuelraj...]

@@markmurto yes I concur with your explanation. Bob Adams pointed this out earlier that this is a special case. Thank you for for clarifying this for me.

[@MrCarburettor]

You are confused because he explained making wrong assumption, this is not related to tangent circle theorem. It is driven from 2 to 4 relationship so FM is parallel to AC as explained here in comments.

[@everybodysvoice388]

Check 14:40 green calculation, 2nd line should = 6.828. Stick to your draft. Thanks.

[@rangaswamyks8287]

I searched for easier methoud than yours.. But i couldnt.. Finally i agread your method is good and correct.. Any way i thank you for solving such fascinative.. Beautiful problem.. Really you are genius.. I like you sir

[@xarran]

I thought I would be using calculus and linear algebra, took me a while to see that Pythagorean theorem would just do.

[@learning_RyanAndArAr]

A very complicated solution...is there any simple way? How did you know that FM=AB+BC even if you have not shown thay FM is parallel to AC or FA is parallel to MC

[@specialcarecars]

Did you get the answer?

[@chilledvibessa5386]

well your answer is correct but your calculation of (4+2sqr2)pr2 is off by a decimal it is not 466,27 it is 46,267... if your are going to be putting up math problems at least have your basic arithmetic in order... did you actually do this problem your self or just copy the example from a source that has there decimal placement off?

[@rolandtragus7482]

I would simplifiy r=12-8sqr(2) so you don’t need to divide by an approximate value

[@govindparmar5956]

Step 5 is not a generic solution. It is correct only if diameter of red circle is equal to radius of blue circle. Other than this case AC will not be equal to FM.

[@YamadaDesigns]

Before watching the video… does the strategy have anything to do with connecting the centers of the 3 circles to make a triangle with sides 6, 2+x, and 4+x?

[@DRJupe1606]

This can work only when given bigger circle radius is exactly double than the smaller one so that third triangle in explanation is right angle triangle. If thats not the case, then third right angle triangle hypothesis would be incorrect.

[@rth314]

It would still be a right angle at whatever point that FM is a tangent to the smaller circle. But then FM wouldn't be equal in length (or parallel) to AC. He should have pointed out they are equal in length by showing that ACMF is a rectangle.

[@rogerio067072]

What got me by the ankles here was that i thought i was assuming something that i could not prove. I dont know why but somehow i saw a rectangular relation but then i thought i was crazy. But i was not. Not yet.🤣🤣

[@ramanivenkata3161]

Excellent working. Well derived.

[@PreMath]

Thanks Ramani for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@gravelordgaming2710]

Nice method : when all three circles have a common tangent and all circle touch each other Then radius of smaller middle circle is given by 1/sqrt (radius of middle smaller ) =1/sq root (radius of circle 1 ) +1/sqroot (radius of circle 3 ) By this formula answer will be 12-8root2 that is matching with answer given in video upto 2 decimal points

[@JVsMusicalSoundscapes]

I guess that's what the 3 equations in my comment above should reduce to

[@philipkudrna5643]

I arrived at r=4(3-2sqrt(2)) or 0.6862 and was very unsure, if this was the correct result. But as it turns out - it was!

[@JVsMusicalSoundscapes]

I solved this using the same approach, but using variables instead of constants (using the symbols in your figure at e.g. 13:53) let rN be the known radius of the red circle let rM be the known radius of the blue circle let r be the unknown radius of the smallest circle (your label) let x be the distance BC let y be the distance AB You get 3 equations in 3 unknowns (x, y, r): (rM + r)^2 = (rM - r)^2 + x^2 (rN + r)^2 = (rN - r)^2 + y^2 (rM + rN)^2 = (x + y)^2 + (rM - RN)^2 with some tedious algebra (which I'm lazy to do), you should be able to derive an expression for r in terms of rN and rM

[@otvirani]

Beautiful!

[@jayeshkumar3861]

It's two step Question if written on paper. DT length = sq rt of [(4+2) square - (4-2) square] which is sum of 2 DTs = sq rt of [(2+r) square - (2-r) square] + sq rt of [(4+r) square - (4-r) square] Means √32 = √8r + √16r √r = √32 / √8•(√2+1) = 2/(√2+1) = 2(√2-1) => r = 4(3-2√2) = 0.68 Approx

[@jayeshkumar3861]

U used the same method. I just noticed later.

[@BlackOps2543x]

Yeah so I calculated this by going over to desmos and just using their geometry tool because I was not in the mood to do some geometric solving today thank you very much

[@isobar5857]

The brain of a professor...and the patience of a saint.

[@CristianNShadowCast]

I was taught to only substitute at the end. r=R1R2/(R1+R2+2*sqrt(R1R2)). When R1=2, R2=4 => r~=0.686 There is a bit of the full demonstration that is left out (if we are pedantic) and that is the touching point of 2 circles is on the line uniting the centers of the circles. Because the radii are perpendicular to the same tangent.

[@ROHITGAUTAM-mv1ws]

How FM become parallel to AC it isn't mentioned in question, moreover 1st circle could be a bit larger or smaller,

[@Muslim_011]

It is. Because the radius 2 is the half of the 4

[@ROHITGAUTAM-mv1ws]

Oh I see thanks

[@Muslim_011]

@@ROHITGAUTAM-mv1ws You're welcome

[@oskjan1]

Yes, that was... exciting. - Raymond Holt

[@wuchinren]

It is an old classic question. Let a is the radius of the left circle, b is the radius of the right circle and r is the radius of the circle we want, then we have identity that √(ar)+√(br)=√(ab) so when (a,b)=(2,4), r=12-8√2.

[@frankchen7436]

there is a formula for this kind of problem, three circles with radius a, b, c, where c is the smallest circle in the middle. Then, 1/sqrt(c)=1/sqrt(a)+1/sqrt(b)

[@murdock5537]

He should have mentioned it - it is more entertaining than the seventeenth example of Pythagorean Theorem...🙂

[@joelhirsbrunner1997]

Or in short, r = (Ra * Rb) / (sqrt(Ra) + sqrt(Rb))^{2}, this is the radius of the biggest possible circle that fits in between two touching circles and their common tangent(s)... Since this is symmetrical, it is the same for either sides

[@andremouss2536]

At last someone who attacks the general problem ! I take you got that beginning with the easier problem of finding the distance between the two tangent points in the simpler case of two circles, then applying the result three times in the original diagram (that's what I did, I think that's the shortest way).

[@ellun3981]

not only the error of 466.27 but he complicates the solution when he can use: 2√2√r + 4√r = FM=√32. it's the same but easier to calculate

[@faisal.k01]

Thank you sir, have nice day❤

[@johncirillo9544]

~466.27 is incorrect. It’s best to work with the radicals, as is. The precise answer contains a radical. Using a calculator to estimate radicals not only loses precision, but introduces the possibility of the type of error you’ve committed. That error aside, the video demonstration is well done.

[@quangphan12345]

mistake on step 5&6 cuz we can't prove that A, N , P on the same line. In step5, we should draw a 90• line to MC instead.

[@MrCarburettor]

Totally agree, he drive that from partial theorem with wrong assumption.

[@shaposhnikovviktor8243]

Does this method work only if the radius of the blue circle is exactly 2 times bigger than the radius of the red one? Because otherwise the NFM triangle will not be rectangular.

[@vmsouzaa]

i think it will be rectangular but FM wont be // to AC. Messing it all up. Am I wrong?

[@professoraman3452]

Salaam... I like all your math videos now I want a perfect solution for this (4-5)^2-2x4x8 = (6-5)^2-2x6x5 And this 1+1 = 3 with proper and valid math applied or boolean calculation ...hope to find prefectly valid solution

[@arthkulkarni5306]

Nice example,best explained.Thanku,sir.

[@SonuSharma-gu1rr]

The way you explain the solution is awesome. I think you are the best math teacher.

[@lonelyorbit580]

How do you know point F is perpendicular to point M?

[@strongma5441]

A LITTLE COMPLICATED. THE DISTANCE BETWEEN TWO CIRCLE IS DOUBLE THE RADIS PRODUCT OF THE SQURE ROOT.

[@pielichiewicz]

15:29 Well - 32/466,27 isn't 0,686 but 0,0686. Earlier 15:11 - if square root of 2 is 1,414, then 2 times 1,414 is 2,828, so 4 plus 2,828 is 6,828, not 68,28.

[@ae00505]

Thanks sir. At one point of time I was getting confused that if right side is 4*sqrt(r) then left side will be 2*sqrt(r) thinking it has a linear relationship with the radius of circles. But that was not true. Then I watched till the end and thanks for the video.

[@carloscarrillo6493]

Yes I found it....! It's right there in the middle of the yellow circle...! You're welcome...!

[@davidseed2939]

Establish the lemma that the horizontal distance between the contact points of two circles with radii a,b is 2sqrt(ab) So 2sqrt(2×4)=2sqrt(2r)+2sqrt(4r) Hence sqrt(8) =sqrt(r)×(sqrt(2)+2) Squaring 8=r×(6+4sqrt2) 4=r(3+2sqrt2) Multiplying by the congugate. 4(3-2sqrt2)= r(3+2sqrt2)(3-2sqrt2)= r(9-8)=r Ie 4(3-2sqrt2)=r r=4(3-2.8284)=4(0.1716)= r=0.686(4)

[@klmkt4339]

I enjoyed walking along the solution

[@mariodistefano2973]

Thanks sir for the explanation! A question: we currently use the square meter as a measure of a generic area, which is based on the assumption of a square with 1m at sides. Hence the surface of a generic circle is function of PI. Now, if we say that a circle of 1 Mt diameter has a surface of 1 ROUND meter, using this strange unit, what would the equivalent PI value, if we use it to calculate in round meters the surface of the above 1 Mt square? Is it possible to solve this problem in your opinion?

[@TheDGomezzi]

This problem is quite simple. In your definition, a round meter is simply pi/2 regular meters. Therefore, the sides of the square will be the reverse of that, 2/pi round meters. We now square this number to get the area, 4/pi^2 round meters.

[@galfawker339]

You could make it clearer that as NF is 2, length of AN is 2 and length of CM is 4, therefore angle NFM is 90 degrees

[@tehatte]

Oh I just made a comment myself on this and didn’t see your comment. Yeah I found the flaw in his explanation about FM as the tangent line.

[@MrCarburettor]

This would not be clearer this is THE explanation. The theorem used is not applicable here.

[@darthboxOriginal]

Thanks, that was really interesting!

[@JarppaGuru]

14.42 i get 46.627 not 466.27.. 4+2*sqrt(2)=6.628 not 66.28 is that some magic something else 2square(2) its not same as 2*square(2) ? if it is you calculate wrong but prox correct thing lol