Are You Ready To Work Your Brain To The Max? | This Is Tricky If You Do not Apply All Your Senses

Added: 2026-05-13

[00:00:00]In this video, they want us to find the radius of the small circle. So, we have this medium circle which has radius of two. Then we have the big circle which has radius of four. And with that, we're required to find the radius of this small circle and the three circles touch at these points. And also there is this line of tangency of which it's a line of tangency on all the three circles.

[00:00:36]Let's start by labeling the centers. Let this be A. Let this be B and this small center C.

[00:00:47]Next let's draw a straight line from point B to the line of tangency.

[00:00:55]Just like this. Also, let's draw another straight line from center C to the line of tangency.

[00:01:02]Just like this. As you know from uh tency to circle theorem, the angle between a tangent or between the radius and a tangent line is 90°. So this is 90°. This is 90°.

[00:01:20]Now let this point of tangency be D and let this point B E. Now as you can see CD is the radius of the small circle and B is the radius of the big circle. Next step is let us connect uh center C to center B [snorts] just like this. And again let's draw a line from center C that is horizontal or that is perpendicular to line to radius B. Okay.

[00:01:56]So here is the line. Now let's call this point F.

[00:02:04]Now let the radius which you want to find of the small circle let it be R. So CD is R. Now because this is a rectangle that we formed here C F E D is a rectangle. So CD is equals to F. So if CD is R then also F E is R.

[00:02:27]Okay. Also CF is equals to D.

[00:02:32]Remember this line here was perpendicular to this line. So we have a right angle triangle CFB.

[00:02:42]So if we focus on triangle CFB, we know that from here to here is the radius of the big circle which is four. So here to here is four and here C to this point is radius R. Therefore CB is equals to 4 + R.

[00:03:06]Also B E is equals to uh the radius of this big circle. So BE is 4 and we know that F is R. Therefore to find just BF BF will be equals to the whole B which is 4 minus F which is R.

[00:03:29]Okay. So now since this is a right angle triangle, this is B, this is C, this is F and this C CB is 4 + R and BF is 4 - R. Now we can find CF using Pythog theorem. So A 2 + B² = C².

[00:03:59]So a is cf. So we shall have cf 2 plus b is this fb which is 4 - 3. So 4 - sorry 4 - r² is equ= to c which is 4 + r this is squared.

[00:04:18]So now we make CF the subject you shall have CF² being = 4 + R 2 - 4 - R².

[00:04:30]Now from a + b 2 being = a 2 + 2 a + b 2 also a - b 2 = a 2 - 2 a + b 2. So using this theory our CF squar will now be equals to so expanding this in this form to be 16 + 8 R + R² then minus expanding this in this form we shall have 16 - 8 R + R² so we open the bracket here we shall of CF² being = to 16 + 8 R + R 2 - 16. This one becomes + 8 R. This becomes - R 2.

[00:05:33]So this R 2 cancels with this R 2 and also have 6 + 16 - 16 cancel. So we are left with CF 2 being= to 8 R + 8 R and so CF² is equals to 16 R.

[00:05:54]So now we find the square root on both sides.

[00:05:58]This cancels. So CF is equals to square root of 16 is 4. So we shall have 4<unk> R.

[00:06:07]Therefore CF is 4<unk> R. But remember CF is equals to D. Therefore DE is equals to 4<unk> R.

[00:06:20]So next step is let us connect center A to center C just like this. So next let's draw a perpendicular line from point A to the line or to the point of tangency just like this. So let me call this point G.

[00:06:40]So this is 90° from tangent circle theorem. Next let's draw a perpendicular line from point C to line AG.

[00:06:53]So I have drawn the line from point C to line AG and it's a perpendicular to AG.

[00:07:00]Let me call this point H.

[00:07:05]Now since CD is the radius of the small circle so HG is the radius sorry HG is R.

[00:07:15]Now since from here to here is the radius of this medium circle and it is 2 and also here to here is the radius of the small circle and it is R. That implies that AC c is equals to 2 + r. Okay.

[00:07:37]And also ag is 2 which is the radius of this circle and from here to here is r.

[00:07:44]So getting a h a h will be equals to 2 - r.

[00:07:52]So now focusing on triangle A H C.

[00:07:57]Okay. So it's a right angle triangle.

[00:08:02]We said this side this is A. This is H and this is C. And A H we said is 2 - R.

[00:08:13]A C we said is 2 + R. So our task is to find HC.

[00:08:20]And since [clears throat] it's a right angle triangle, we're going to use Pythagoras theorem. So a² + b= c². Our a is h c. So h c 2 + our b is 2 - r. This is squ= to c which is 2 + r.

[00:08:38]So 2 + r squared. So we make h c the subject. So we shall have h c^² being = 2 + r² - 2 - r squared.

[00:08:55]So again from a + b 2 being = to a 2 + 2 a + b 2 and also a - b² = a 2 - 2 a + b squared. So we shall have h c^² being equals to 4 + 4 r + r 2 - 4 - 4 r + r 2. So when we open this bracket we shall have h c 2 = 4 + 4 r + r 2 - 4 + 4 r - r² this r² cancels with this r² and also we have + 4 - 4 they cancel so we left with h c 2 being = 4 r + 4 r so h c 2 is = to 8 r. Find the square root on both sides. So h c is equ= to 8 is the same as 4 * 2 r. So h c is = roo<unk> 4 is 2. So 2 <unk>2 r.

[00:10:27]So this is the value of hc. So remember hc is equals to gd. So if HC is 2 <unk>2 R also G D is 2 <unk>2 R.

[00:10:44]Next step is let us connect point A to point B or center A to center B just like this. Let us also draw a straight line from center A to the point to the circumference.

[00:11:01]Straight line to make a diameter G up to a certain point.

[00:11:08]And let us call this point J.

[00:11:14]Now let's draw a straight line from point J to center B.

[00:11:20]Just like this. has realized this is 90°.

[00:11:25]Okay.

[00:11:30]Now since from here to here is radius of this one which is 2 and from here to here is radius of this one which is 4.

[00:11:38]That implies that a is equals to 2 + 4 which is 6.

[00:11:47]Now since this is 90°, this is 90°, this is 90°, this implies that J B E G is a rectangle. So implying that G E is equals to J B. You'll forgive me the diagram is not to scale, but that is what it is. So G E is equals to JB.

[00:12:14]So JB will be equals to this GD which is 2 2 R<unk>2 R then plus this 4<unk> R.

[00:12:30]So we factor out roo<unk> R.

[00:12:34]So JB will be equals to<unk> R into this will be now 2<unk>2 + 4.

[00:12:53]So this one will be equals to<unk> R into 2<unk>2 + 4.

[00:13:03]So now considering triangle J A which is a right angle triangle. So this is J B A and this A is 6. This J A is 2 and this is roo<unk> R into 2<unk>2 + 4.

[00:13:27]So using Pythagoras theorem which is this is 90°. So this is a squ + b² = c².

[00:13:39]Our a is jb which is so it will be<unk> r into 2<unk>2 + 4. this all 2 + 2 = 2 a which is 6 squared.

[00:14:02]So now this here will become<unk> r² into 2<unk>2 + 4 2 + 2 2 = 6 2. This cancels. So we left with r into this 2<unk>2 + 4 squar + 2 ^ 2 is 4 = 6 2 is 36.

[00:14:32]So shall have r. So this one we shall will be 2<unk>2 + 4 into 2<unk>2 + 4. then this will be equals to 36 minus 4 when this crosses.

[00:14:48]Okay. So here we're going to use distributive to multiply multiply this by this and this by this then multiply this by this and this by this. So we shall have r into 2.

[00:15:07]So 2<unk>2 * 2<unk>2 gives us 8 because 2 * 2 is 4. Then <unk>2 * <unk>2 is 2. Then 2 * 4 which is 8. So we shall have 8. Then plus 2<unk>2 * 4 gives us 8 <unk>2. Then + 4 * 2<unk>2 is 8 <unk>2. then + 4 * 4 is 16.

[00:15:34]This is equals to 32 - 36 - 4 is 32.

[00:15:41]So we shall have r into 8 + 16 gives us um 30 sorry 24 then 24 + 8 <unk>2 + 8 <unk>2 gives us 16 <unk>2.

[00:15:59]Okay, this is equals to 32.

[00:16:07]So this will be r into 16<unk> 2 gives us 22 something. So when you add that to 24 you get 26.62 74.

[00:16:24]Okay.

[00:16:27]So this is equals to 32.

[00:16:30]Therefore we want to make r the subject.

[00:16:32]So we shall divide by 26.6274 by 26.

[00:16:38]6274.

[00:16:42]This cancels and therefore r which is the radius of the small one that we are looking for will be equals to when you divide this we shall have 1.21. 2 018.

[00:16:59]So this is the radius of this small circle. It's 1.202.

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