This circle geometry math question will challenge you.

Welcome back to my channel. We have a challenging geometry math question to solve. In this question, we have two circles. Please, this is not drawn to scale. So we have two circles, and the diameter of the bigger circle is 14 cm, which means that the radius of the bigger circle is 7 cm. So the question says that we should find the value of Pq. That is the question. Okay. First things first. Join this to this. Let's join Q to the center of so. And then we join that to P. Join to P. Alright. Now, let us name this O. The center of this place O. So now we have triangle PQO or POP. Now, angle P is 90 degrees. And that is because when a tangent meets the radius of a circle, their point of contact is always 90 degrees. When the center of a circle meets the tangent of a circle at its point of contact, they form 90 degrees. Alright. So this place is 90 degrees. Now, let's call OQR, which represents the radius of the smaller circle. So radius of the smaller circle. Alright. Now, using Pythagorean theorem, using Pythagorean theorem, we will have that PQ squared plus QO squared or OQ squared should give us PO squared. Alright. So it means that PQ squared plus R, because OQ is R, R squared is equal to PO squared. Alright. Okay. Now, let us join the center of O to draw a perpendicular from center O to touch the diameter of the bigger circle. So having done that, because it's a perpendicular, this is 90 degrees. And let's call this place A. Let's call this P. I have my reasons. So let's just call this AB. Alright. So OQ becomes BQ, right? So we have BQ here, and we've corrected it here as RO squared. So we are going to write this first statement, that PQ squared plus RO squared will give you PB squared now. So it's going to be PB squared instead of PO squared, right? So it's equal to PB squared. Alright. Now, clean this because at this point we don't really need it anymore. Have you subscribed to my channel yet? If not, please subscribe, turn on your notification bell as well. That way, whenever I upload a new video, you will see it. They will send a notification to you to show that I uploaded a new video. You always know that I'm in the comments, so drop your comments in the comments section. Whatever it is you want to, let JJ know. I will be in the comments reading it, and I will reply as well. Now, looking at this, we have another triangle, triangle PAB, right? So we have PAB, triangle PAB. Now, using the Pythagorean theorem as well, we are going to have that PB squared is equal to PA squared plus AB squared. Now, is there a way that we can actually get a value for PA? Yes, there is a way to do that. Now, remember this is not drawn to scale because I'm already seeing this is going up. I think I need to correct that. So I'll try to see how real I can make this look. Okay, alright. Now, this is 90 degrees here. If you look at A, let's call this point M. Now, let's join this down and let's call this point N. Okay, let's call this point N. Now, BN is the radius of the smaller circle. Therefore, AM is also the radius of the smaller circle. And if PM is 14 cm and AM is R, it means that PA should be equal to 14 minus AM. Yes, so that means that PA will be equal to 14 minus the small r. So, this place is 14 minus r. Therefore, in triangle PAB, we are going to have PB squared is equal to PA squared. 14 minus r squared plus AB squared. Of course, we don't know what AB squared is. Now, remember this equation. Let's call that one and call this two. You can see that both of them can be equated together because both are equal to PB squared. So, this simply means that I can equate this to be equal to this. So, I can write that 14 minus r all squared plus AB squared is equal to PQ squared plus r squared. Okay, so now we have this equation in terms of AB squared and PQ squared. And I don't want to expand it yet. I'll just leave it like this or let me go ahead and expand it. Let me go ahead and expand it. So, I'm going to write this over. PQ squared is PQ squared plus r squared equals PB squared. Okay. Now, so we are going to expand this 14 minus r squared using algebra identity. That says that A minus B all squared is equal to A squared minus 2AB plus B squared. Therefore, we are going to expand this. So, doing that, we have 14 squared which is 196 minus 2 times 14 is 28 r plus r squared. Now, plus AB squared is equal to PQ squared plus r squared. Now, if you look at this, you see that we have r squared there and r squared there. So, both of them we cancel out. So, if that happens, we have PQ squared is equal to 196 minus 28 r plus AB squared. So, let's leave this and clean every other thing for now. It's no longer important. Have you hit that subscription button yet? If you have not, please do that. So, this is all we have now and we are leaving it. Now, I want us to join this center. Now, let's call this center P. This center, let's call it P. The center of the bigger circle. Now, let's join P to B. Okay. Let's join P to B. So, let's join P to B. So, if we do that, we now have another right angle triangle PAB. So, let's draw it out. Triangle PAB. This is P. This is A. And this is B. Oh, we have P already. So, we cannot call this place P. Let's call it O since we don't have O again. So, this is O. So, we have OAB, right? Now, we are going to use a lot of theorems here. Okay. Now, if from O to M. If OM is R plus something. Okay. Let's do it this way. From O to A. It's definitely going to be OM is the capital R. Right? The radius of this circle. So, let's say OM is equal to capital R. That is the radius of the bigger circle. Now, OA. It means that OA will be O2. OM minus AM. Right? Minus AM. Now, OM is the capital R. The big radius. And AM is the smaller radius. You can see that. So, it means that OA is capital R minus small r. So, you have this minus that. For the other side, we have AB is AB. There is nothing we can do about it for now. Now, let's talk about OB. You can see that OB joins the two circles together. Because when two circles meet each other externally. Okay? A straight line joins the two centers and the point of intersection. They all align on a straight line. Yes. It's a theorem. Now, looking at this. So, it means that OB will give us the capital R. Which is from O to this place. And from here to here is the small r. So, OB will be capital R plus the small r. Now, you can see that we can solve this using the Pythagorean theorem. So, using Pythagorean theorem, we will have OB squared. Which is capital R plus small r all squared. Will be equal to OA squared. Which is capital R minus small r all squared. Plus AB squared. Right? Okay. So, I can clean this now. So, that we will have a space too. Now, moving forward. We have. If we expand this. We will have capital R squared plus two capital R small r plus R squared. Is equal to. Here, we will have the same thing. Except that this is negative. So, we will have R squared minus two capital R r plus R squared. Then plus AB squared. Now, let's see if there is something we can cancel out. Oh, thank God. We cancel R squared, R squared. And we cancel this and this. I love doing this when I solve. Because once you cancel this out. Everything becomes smaller. And you are getting closer to your answer. Okay. So, this is what we have left. Plus two R. So, we have two R r. Is equal to minus two R r plus AB squared. So, if we transfer this to this side. We have two R r plus two R r equals AB squared. So, we have four R r equals AB squared. Now, this is telling us that AB squared is four times the capital R and the small r. Now, we already know the radius of the bigger circle. This one. Because we are told that BM being the diameter is 14 cm. So, the radius of the capital R is seven. So, it means that four multiplied by seven is equal to AB squared. So, we have 28 R is AB squared. Permit me to write it somewhere. So, I don't know if you can see it. You can't see it. I know. I'm sorry. Just have to write it somewhere. Because I need to clean this. Now, I know this is getting longer. So, I have to check to be sure it's still recording. Sometimes, because of space, my phone has... The space is filled up already. And I don't know what to delete. The only thing that I'm going to do is to buy a new phone. Yeah. I need to buy a new phone to be able... Something better. So, that I will have space in my phone. Alright. So, now we have PQ squared equals 1, 9, 6, 1, 0, 28 R. Remember this equation. So, remember our 28 R is... Okay. So, now we have AB squared. That is what I cleaned here. Is equal to 28 R. So, we are going to replace this with 28 R. So, if I write that now. We have PQ squared is equal to 1, 9, 6, minus 28 R, plus 28 R. Okay. So, you can see that this and this will give us what? They are using inverse property. So, this and this will give us 0. Now, we have PQ squared is equal to 1, 9, 6. So, we take square root of both sides. PQ is equal to 14. So, 14 units squared. Okay. We have centimeters. And that is the answer. Thank you so much for watching and see you in my next video. Bye.