 We start giving your attendance in the chat box Right wait for two three minutes so that all of you can join and then I will start teaching please Write your names in the chat box Gopika good Anyone else who has joined Shani is there. Okay? Anyone else? Okay, I know Ragh Nihar good so Guys, what do you as this is a revision class? What do you want me to do? Do you want me to do theorems only or do you want me to do questions? Or do you want me to mix everything? Just let me know yes circles only oh, okay both so fine I'm starting with Theorums first and I'm starting from very beginning if required. I'll take two classes of circles doesn't doesn't matter so I'm starting with theorems So I'm doing circles only. Okay. Let me write here so that I'm going to do circles only I'm waiting for two more minutes. I don't want anyone of you to miss this class because We have already done one revision is still there was problem. I'm doing this Can you guys check with others whether they are joining or not? I have only four four people now till now Anura, can you check with Karthikeya? Nihar, can you check with Nishant? Okay, no issues. Okay, so then I'm starting the class now So I'm going theorem by theorem theorem one is and it's about Congruency of circles So what does it tell congruency of circles? so it tells that Two circles are congruent and only if they have equal radius Okay, Anura. I've started the class. So Two circles are congruent if and only if they have equal radius So suppose I have this circle. I name it C1 and this another circle I name it C2 and the radius is R1 and Radius is R2 here if I am saying that circle C1 is congruent to circle C2 It tells me that this is only possible when R1 is equal to R2 So this is this is not a theorem. I'll call it. This is an exam because If radius is equal circle would be overlapping on each other hence I mean if radius of two circles are equal then The two circles would be congruent now. Let me go to theorem two and Theorem two twos nearly theorem eight or nine Nearly theorem nine. So theorem two to theorem nine would be theorems about chords chords of circles so theorem two is Equal chords Substant subtend equal angle on the center of the circle or at the center and vice versa is vice versa is if angle Subtended by two chords. So this is theorem two and three. I'm proving it simultaneously if angle subtended by two chord Cords at Center are equal Cords would be equal. So suppose this is a chord AB and This is my center O and this is called CD. This doesn't look like same but I Will do this construction. So I have two triangles triangle a OB and triangle COD So in first case what will happen is for theorem two I'll I'll I'll wipe it out and and and prove theorem three for theorem theorem two AB is equal to CD It has been given to me and oh a is equal to OD and OB is equal to OC. These two are radius So hence side side side Congruency is there and at side side side congruency is there by CP CT Angle opposite to AB angle opposite. So AB is equal to CD So angle opposite to equal sides would be equal opposite to AB angle a OB is there and Opposite to CD angle COD is there. So angle a OB is equal to angle COD Now what happens in in case you have to prove the other theorem theorem three that if angle Subtended is equal. So there a OB is equal to angle COD would be given and I know that oh a is equal to OD and OB is equal to OC. This doesn't need to be proved This is radius. So now this is side angle side congruency So by CP CT what I can write is Opposite to equal angle sides would be equal. So AB is equal to CD So in one case you have side side side congruency in other case you have side angle side congruency So these were theorems two and three if at any point you guys are not Understanding any question or any theorem. Please write it down in the chat box Now theorem two and theorem three I have already proved Theorem four is actually Actually the same thing it is saying that if two arcs of a circle are congruent Then the corresponding chords are are equal So what I do is I take this circle and I have called AB here and I have called CD here Let me mark it. This is AB and This is CD and this is my center. Oh now. I am saying you that called AB and Sorry arc AB is equal to arc CD Sorry congruent to card CD. So what does it mean? So there are three cases. So case one is that AB and CD are arc AB and CD are Minor arcs So if AB and CD are minor arcs, let me do this construction so congruency tells me that if Two arcs are congruent. This is a geom If two arcs are or you can write theorem because in your book it has been given as theorem if two arcs are congruent They subtend Same angle on the center. So now angle If they are minor arc So I will write that way is equal to OD OB is equal to OC So these are my radius The two triangles you already know they are OAV and OCD and As I'll write angle AOB is equal to angle COD and here in description. I'll write that arc Minor arc AB is equal to or congruent to minor arc CD and hence I'll write here Hence angles would be would be equal So now you have shown that these these angles are equal So side opposite to equal angles would be equal then you can write AB is equal to CD if You have now now in case of major arc If if I in case of minor I write major So I have to write like this that if Minor arc AB is congruent to CD Hence major arc So AB is minor arc. So I'll go I'll write BA major arc BA would Also be congruent to major arc DC and Once you write the angle again become same So if this this angle you assume to be theta the other angle would be 360 minus theta So if this is 360 minus theta, this is theta So you can you can prove by same method that this this would be similar This would be same hence the cards would be same. So there is no other method You just have to one line that if minor arcs AB and CD are same major arcs BA and DC would also be same So in that in that case the Two cards would be same So I have done it for both major and minor arc There is very little difference in minor arc You can directly write that the angles are same in major arc You have to first show that the minor arc is same and then you have to write that the major arc is same now Theorem 5 is exactly the opposite of it and which says that if Cards equal I'm not writing the complete theorem because it is taking a lot of time if chords equal Then arcs are same Arcs are congruent So you don't have to do anything. I have already told you that if are if if AB is one chord and and CD is another chord and I do this construction center is o I have in triangle OAB and Triangle OCD I have AB is equal to CD it has been given to me and OA is equal to OD and OB is equal to OC because they are radius so side side side congruency and Hence by CPCT you have angle AOB equal to angle COD and When these two angles are equal So if angles are equal I can write that minor arc AB is equal to minor arc CD and Once you write this You can say that AB is congruent to CD Now if AB is congruent to CD so this is for minor arc for major arc what we need to do is that We need to write that That as AB is equal to CD so VA would be equal to DC You just need to add and this minor Major arc would be equal now you need to write that If this angle is theta the other angle would be 360 minus theta here also it is 360 minus theta It shows that this these major arcs are same Now if these this if this is making angle theta these two arcs would make angle theta and you need to write that major arc BA Would be congruent to major arc DC and once you write that you can write that the chords are equal or if chords are equal This this possibility is there in case of semicircle you can write that the two arcs are same They are making 180 degree angle So hence the two chords would be Same is semicircles to semicircles of a circle are congruent to each other Now another type of the the all all these were from theorem 2 to theorem 5 where all same What on a honor the which question you are asking? BDN AC I am not saying BDN AC are same so The next theorem is okay up you you can ask me ask that question to me later Individually, I'll clarify it this I am going to theorem 6 and Theorem 6 tells me that the perpendicular drawn from the center of the circle Bisex the circle so perpendicular to chord Bisex it to chord from center. I'm writing here not from any other point From center bisect it. So how do I do it? I know that I Will suppose this is oh, this is a this is B and this is C. So I have two triangles OCA and Triangles OC B and What I'll do is I'll write that OA is equal to OB OC and OC is common and Angle OCA is equal to angle OC B is equal to 90 degree it has been given to me this is common and This is radius Hence I have RHS congruency and As I have RHS congruency. So side opposite to equal angle opposite to equal sides would be equal So opposite to OA the angle is OCA and Opposite to OB it is angle OCB So angle OCA is equal to OCB. I also know that angle OCA plus angle OCB is Equal to 180 degree and as they two are equal suppose they are equal to X So 2 X is equal to 180 X is equal to 90 degrees So I can write that this angle is 90 degree and this is how I prove it now Converse of this theorem tells me that the line drawn through the center of a circle to bisect the chord is perpendicular so In in in converse of this theorem what happens is I have been given that Here one thing was left out is that that I have to prove that it bisects is so this angle and this angle because What happens is in in what I proved was AC is equal to BC and opposite to Or you can write that this angle is equal to this angle. So angle AOC is equal to angle BOC and If that is true, then AC becomes equal to BC So you can say that this has been and and converse of this theorem would be that you have to prove that it is Perpendicular so you will be given that this bisects this so if From center I draw a line the theorem is this is the center of the circle This is called AB and drawing a line which is bisecting the card So I am I'm doing this construction way and OB so I write that way is equal to OB I have been given that AC is equal to BC. This has been given to me and Then I know that OC is equal to OC. So side side side congruency hence This angle would be equal to this angle and by adding it you can prove that this is 90 degree, which I just did So these two theorems get proved easily now Let me tell you I have to prove that There is a corollary which needs to be proved And this is very important So look at it The perpendicular bisectors The perpendicular two chords Of a circle intersect at its center. So what I have to do is I am drawing the circle And this is one of my chord and this is second of my chord I give its name AB and CD And I have a perpendicular bisector So suppose this is E and this is point O And I have another perpendicular bisector which is OF And suppose instead of O This perpendicular bisectors meet at ODAS So why I'm drawing ODAS I'm explaining here I am drawing ODAS so that I can prove that O and ODAS coincide If I'm able to prove that O and ODAS coincide in that situation what happens that I'll be able to prove that any other point other than O Taken as the point of intersection of perpendicular bisectors of two chords Would meet at O only any other point if I take and and that point coincides with O That will prove my theorem. So I have to prove here that Prove that ODAS Go insides with So For that I have masked OE and OF Now there it has been given that they are perpendicular bisectors So proof is E is midpoint of AB now I just told you that if E is midpoint of AB and OE is perpendicular to AB And ODAS E also is perpendicular to AB. I have taken two points O and ODAS This is only possible. This this ODAS E This is only possible When ODAS E Lies along OE Similarly, I can write that OF is perpendicular to CD And ODAS F also is perpendicular to CD and this is only possible when ODAS F lies along CD So Sorry OF So What happens is I have proved that ODAS E lies along OE and ODAS F Okay, so Anurag what I'm trying to do is in this in in this in in this corollary is It has been asked to me to prove that whatever perpendicular bisectors you we draw on cod Passes through center. So the idea behind this theorem is That I have to prove that all the perpendicular bisectors of I mean any chord will pass through center How will I prove that? Any perpendicular bisector will pass through center. So for that what I'm doing instead of center I am taking any other point ODAS and I am Joining ODAS with the midpoint of the chord. So I have joined ODAS with E. I have joined ODAS with F And I have center O. So I have joined OE also and OF also From the previous theorems, which I have proved. I know that the perpendicular bisector OE Has to pass through center. So I am writing that OE is perpendicular to AB But what I have done done is I have taken any other point ODAS in the circle And I'm I'm saying that ODAS E is also perpendicular to AB. Why ODAS E is perpendicular to AB? Because I have assumed that ODAS is such a point that ODAS E is perpendicular to AB. So this I am writing by previous theorems. So this Is on the basis of previous theorems, which I just proved And this I have assumed that any other point old ODAS is taken And this I have assumed So these both these cases will not be possible. This will only be possible when ODAS E and OE would be a same line So that is why I have written that ODAS E lies along OE and similarly for other case also Now, did you understand? Did you understand Anurag? Anurag, are you there? Okay, fine. Anurag perhaps is not here. So ODAS E And ODAS F lies along OE and OF respectively So it means that intersection of point or of intersection of ODAS E and ODAS F Would be ODAS and ODAS coincides with O So that's how you need to prove it Now let me move to another Theorem which is Now theorem 8 tells me that equal chords equidistant From each other. I draw a circle like this and I have a chord AB I have another chord CD Now distance of please understand this Distance of support. This is point O distance of O From any line or distance of any line from a from a point is Is always taken as a perpendicular drawn from that point to that line So if I have a point here and I have a line here I will not take this distance as distance from suppose this is AB and this is C So distance between C and AB is not this distance distance between AB and C is this perpendicular distance So CD is distance between point C And line AB Now here I will match these two So what happens is I have to prove that I I draw a perpendicular here. I have to prove that This is not D. Suppose. This is m and this is m So I have to prove that om is equal to on So I have been given AB is equal to CD and I know that Or what I do is I take these these two triangles. So I take triangle o a n And triangle o c m These two triangles triangle one and triangle two. I'm taking so here. I have been given that I know that in this triangle o a is equal to o c because this is radius r and I know half of AB because AB is equal to CD So half of AB would be equal to half of CD And half of AB is equal to half of CD which says me that a n is equal to c m and I know that This is 90 degree as I'm drawing perpendicular from here. So That would be nothing but o n a is equal to angle omc Which is equal to 90 degree So it means that triangle o m c and triangle o n o n a Are congruent to each other if they are congruent to each other it means that These two angles are equal or by cpct you can write that om is equal to om so AB and CD are equidistance distant from o now Opposite of it. So it has been saying that cards which are equal equidistance from There are too many theorems in the beginning. There are 18 theorems. So it will take some time for me to finish off these theorems so Cards Nehal just give me 10 minutes. I'll start solving questions Cards equidistant from each other Not each other. Sorry equidistance from center are equal So suppose this is AB and this is CD So this is a perpendicular drawn This is a perpendicular drawn is o So what happens here is I need to prove that And this is suppose m and this is n So I need to prove that AB is equal to cd This is to prove Now, how do I prove it? So let's take triangle. I know that AM is equal to half of AB and cn is equal to half of CD So if I am able to prove that am is equal to cn Then I am able to prove that ab is equal to cd So let me write here I'm taking triangles om a And triangles o nc So oa is equal to oc and I have been given that Om is equal to on because they are equidistance angle om a Is equal to angle om c which is 90 degree Hence these two triangles are congruent and once they are congruent you can show that am is equal to cn You multiply two to it. You will write ab is equal to cd Now what I'm doing is I'm on on the basis of this theorem is only which is there in your book You solve two theorems. You will understand it completely that one theorem is Of any two chords Which is nearer to center The one which is of any two cards the one Which is nearer to center will be Longer And one which is farther for from center would be smaller. So you you can take this theorem. You can look at look look in the book And oppose it of this So now I'm taking a few questions on the basis. There are a few theorems for congruency of Two theorems. There is nothing in that just go through from Congruency of two circles. Just go through the theorems in your book. You will understand everything Congruency of two circles You will understand everything every time you will have to prove that the radius is equal and obvious of Obvious things would be given there. So These were few important The theorems which I took now. Let me solve a few questions from for you The first question which I'm taking is Write down this question two chords ab and ac of a circle are equal prove that the center of the circle lies on the angular bisector of angle b ac Just solve this question I give you two three minutes to solve this question And let me know when you are done with the question Hello Yes, who is doing Okay, ishani has done it Obviously, you have to do the construction on Iraq without construction. How will you solve it? I'm waiting for others Okay, Nishant got it two more minutes Okay, so all of you got it Anurag, I'm just waiting for you if you're not getting it. Just let me know Okay, so as Gopika is telling me to explain I'll explain it So how to do this question? It's like this So what we need to do over here is there are two chords ab and ac So this is my circle and This is ab and this is Or let me make another chord. This is ac And this is point So and this is Any points for this point? Oh And this is d Let me match this And let me call it m so suppose ad is Bisector of angle b ac ad one second suppose ad is bisector of angle ad is bisector of angle b ac so What happens is I have to prove that prove that oa lies on ad This I have to prove that oa lies on ad So for that what I'm doing is I'm constructing a line where where I'm drawing bc Now I take triangle b a m And triangle c a m So I take these two triangles. I know that ab is equal to ac because it has been given to me like this Angle b a m is equal to angle cam because The angular bisector as a o is angular bisector So angular bisector will divide the angles equally. So hence these two angles are equal And here a m is equal to a m So I have side angle side congruency And inside angle side congruency. I have b m is equal to c m and I also know that These two angles would also also be equal because these two angles are opposite equal side b a So b a and ab and ac so I know that b m a Is equal to angle c m a And I know that angle b m a plus angle c m a Is equal to 180 degree Hence the both would be equal to 90 degree. It says that a m is is perpendicular bisector of Am is perpendicular bisector of bc and if If if it is perpendicular bisector of bc So you try to understand after this it's only about writing Uh adequate things and how do you write it? So you write that Am is part of ad So it means that a ad ad is perpendicular to bc or perpendicular bisector to bc of bc And if ad is perpendicular bisector of bc I know that all the perpendicular bisector. I just proved the corollary I I wrote that all the perpendicular bisectors of a chord will pass through center. So it means that o m has to be part of Ad and if o m has to be path on part of ad it means that o lies on ad So that's how you have to prove this Any doubt now? Okay Now let me give you another good good quality questions so the question is if Two circles intersect In two points prove that The line through their centers Is perpendicular bisector of the common chord Just prove it Anurag if you have done it differently the previous question just show it to me in the class I'll check whether it is correct or not Okay ishani got it Have you done how many of you have done? Only Gopika and ishani has done. Okay. Let me solve it enough time Nihar got it once again started looking This is one circle and this is another circle So what happens here is this is A line this is o this is O dash I draw a line from o to o dash And then I draw a line here. This is a and this is b And this is radius r and this is radius s so I have to prove that O o dash is perpendicular to a b. This is what I have to prove And if this is what I have to prove the best way to prove is to prove that angle o a So suppose this point is m So angle o m a is equal to angle o m b if I'm able to prove it I know that Addition of o m a and o m b is 180 degree and both are equal. So hence that would be 90 degree So what I do is I take triangle o a o dash and Triangle o b o dash and I'll prove that they are congruent So what I do is I write o a is equal to o b here. I write O dash s is o dash a is equal to O dash b. This is radius r. This is radius s and O o dash is common to both of them This is common. So hence side side side congruence is there And if side side side congruence is there Then I can write write down two things So which is which is this that this total angle o angle o A o dash is equal to angle o b o dash This is equal by cpct I can also write that angle o Sorry angle a o O dash is equal to angle b o o dash This also I can write So what happens is Now look at here now this side And I can also write that A o o dash I have written so it means that this angle is equal to This angle and if this angle is equal to this angle Then I can write that angle A O m which I have already written here that angle A o o dash so I can write here angle a o m is equal to angle Uh b o m So this is nothing but this only just instead of o dash And m falls on o dash. So what What I can do is I can write from here that Now let me take triangle o a m and triangle o b m and in this triangle, I know that oa is equal to os I also know that o m is common to both of them And I have already proven that angle A o m is equal to angle b o m. So I have Side angle side congruency and from that I can prove that Angle by cpct. I can write here. I'm writing here by cpct I can write here that angle o m a is equal to angle o m b And they both are equal and their addition. I have already told you here that Their addition would be equal to 180. So you can write that this is equal to 90 degree As simple Now let me give you another question And the question is write down the question. I'm orating the question write down the question Let's not waste the time. I'll write it prove that the perpendicular bisector of the cord Prove that the perpendicular bisector of the cord Of a circle or always passes through center Over leave this question. We have already done this. I thought this is a different question One second This perhaps I have done in the leave that question Which I was trying to give you This perhaps I have already solved in the So I have two cords ab and cd ab is equal to cd And center is o And radius is r So what happens here is this these two cords intersect at e And from o I draw perpendicular So this is 90 degrees. Suppose this is m. This is n This is also 90 degree So you have to prove that o m En Is a square Yes ab is equal to cd. I have written O is the center of the circle and ab is equal to cd Okay, so one person has got it Gopika got it. I'm waiting for others Anyone who is not getting the answer Nisham got it ishani got it I'm waiting for alura and nihar I think karpike has left Okay so Let me solve it for you guys one more question and then I'll I'll wrap up For the question this question can be solved like this So Join this oe So that you can form two circles and I know that If these are midpoints, so I've already given you this is perpendicular So you know that o m is equal to o n and O e is equal to o e and angle o m e is equal to angle o n e So by r h s congruency, sorry Not r h s. What is this? You can also write that Yes, r h s only what what's more r h s congruency You can write that m e is equal to En I already know that Try to understand I already know that o m is equal to o n because the cards are equal hence perpendicular bisectors on it would be equal So all four sides have shown equal Now I need to show that This is already 90 degree. So this and this is already 90 degree So two angles are 90 degree. What about other two angles this and this? so this Doesn't need to be proved. You can write that if these two angles are 90 degree o m e n would be a square That's how you you get answer of this question I'm I'm giving a next question Which is show that if Cards of a circle Bisect each other There must be diameters of the circle Solve it Let me know when you guys are done Nishant got it anyone else Ishaani got it The one moment when I solve Okay, let me solve it so Suppose I have a Circle once and let me make a good circle. My drawing is very bad I have a circle like this and I have called a b and I have called c d so What it is here is this is o And they are intersecting at point o And they are bisecting each other it has been given so I have been I'll have to take o b is equal to o a is equal to o b and o c is equal to o d That has been given to me Now to prove this I'll I'll I'll join all these points So what I do over here is That I take Triangle a o c and triangle opposite triangle b o d And what I get over here is that o a is equal to o b and o c is equal to o d And angle a o c is equal to angle b o d Why because because this is vertically opposite angle So it means that triangle a o c is congruent with triangle b o d And in this particular case A c is equal to b d Similarly, I can take triangle a o d and triangle b o c And I can write that O a is equal to Triangle a o d and leo c so what I can take here is So a o is equal to o b And o c is equal to o d and I can also write that angle a o d Is equal to angle b o c So this is vertically opposite angle So I can write that a d is equal to b c But what do I get by writing this so as soon as I write this I also know that This Minus segment a ac Is equal to Dd and here I get Segment ad is equal to segment bc So what I get is I can add these two. I know ac plus ad Is equal to bc Plus bd Now suppose this total is x and this total is x So ac plus ad Plus bc Plus bd Is is complete circle. So it's circumference of the circle. So it means that this is semicircle And this is semicircle So If these two are semicircle, so It it tells me that Ac ac plus ad is nothing, but you can write c ad is a semicircle equal to The other part of semicircle, which is c vd And if these two are semicircle, then you can write that cd is Diameter And similarly you can write that from this side ac plus bc you can write that write it And and ad plus bd you can write and you can show that ac is a diameter So that's how you can show that These two are diameter to each other So now what I say is That these two are diameters so I'm wrapping up the class now because it's time On Wednesday, I'll give you an envelope test which will be best on quadrilaterals and circles as I have announced in the class So I would like all of you to Um Just focus on that test and get good marks in that test You should submit the test in in in friday's class So Be ready for the test So thank you so much for joining the class See you on wednesday. Thank you. Bye. Bye