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My method was to take b as origin and find center of circle knowing that the circle passes through (0,6) and (2,0) We get a quadratic but we can neglect one value knowing that the center of circle lies in 2nd quadrant Using equations for geometry makes the job a hell lot easier And the ob is basically dist of center from origin

The issue with many teachers is that before starting cordinate geometry they don't teach euclid's / that basic geometry that's why question like these seems impossible

It becomes simple when you take obas origin A(√50 cosx,√50sinx) Thus C(√50cosx + 2,√50sinx -6) are also on the circle, get the ratios and answer what ever you want

Trying before solution: got √26 ,method plotting a rough sketch and brut forcing,circle of radius √50 with centre at origin, trying different chords by taking lines parallel to y axis x=1,2,3,4,5 and calculated distance of P(point of intersection that chord and x axis )and point Q (point of intersection of circle and x axis) which nearly slightly greater than 2 ,then tried to find B on line x=5,by taking various lines y=-1..,got Point B in first try ,so now it was easy peasy just applied Pythagoras,ob²=op²+pb² ,so got 25²+1²=26 hence √26

it can be solved easily by taking o as origin and A as (h,k) , B as (h,6-k) ans C as (h+2,6-k). And as A and C lies on circle X^2+Y^2=50, we can fiind h and k and then we got the point B as (5,1) and the distance between originf and B can be find easily using distance formula which is root 26

sir i have a very simple solution,, extend AB to meet the circle say at D let BD=x similarly extend CB to meet the circle at E Using POWER OF POINT AB.BD=CB.BE we get BE=3x Drop perpendicular from center to BE and AB at F and G we get OF=3-x/2 and OG=3x/2-1 Using pythagoras we get x=4 THEREFORE OB^2=OF^2+OG^2 OB=sqrt(26) !!!!

Here's a solution using "Power of a Point". Extend line AB to meet the circle again at D. By Pythagoras on ABC we have AC=sqrt40. Let angle ADC = x. Then, by sine rule, we have sin(x) = AC/2r = 1/sqrt5, which means tanx = 1/2 = BC/BD = 2/BD which gives BD = 4. Hence, |Power of B w.r.t. circle| = |BA*BD| = |r^2-OB^2| = 6*4 = 24 = |50-OB^2| and hence OB^2 = 26. In general, the answer is r^2 - BC^2 *sqrt(4r^2-AC^2)/AC

I think there is no need to make it complex Method 1: By extending AB let it meets the circle at point E and by extending CB it meets the circle at point F Join point A and C Draw OD perpendicular bisector of the chord AC , meet O with point A AO=R, AC =2√10 and AD=√10 Let angle AOD =x , tanx=1/2 (in triangle AOD) Draw CE then same chord AC will construct angle x (1/2 of the angle made at the centre) at point E Now in triangle BEC Angle BEC=x And tanx=1/2=BC/BE=2/BE BE=4 AB×BE=BC×BF BF=6×4/2=12 Draw OP perpendicular to AE and OQ perpendicular to CF We get OP=6-5=1 and OQ=12-14/2=5 OB²=OP²+OQ²=1²+5²=26 Method 2: Let the coordinates of point B : (0,0) and coordinates of the centre of the circle be: (-g, -f) Where g is +ve and f is -ve OB²=g²+f² the equation of the family of circles passing through the points (2,0) and (0,6) would be x(x-2)+y(y-6)+k(3x+y-6)=0 g=(3k-2)/2 , f=(k-6)/2 , c=-6k R=√(g²+f²-c) by putting value and then solve we get k=4 ✓ g=5 , f=-1 OB=√26

Produce AB to meet the circle at X and Produce CB to meet the circle at Y. Let BX = a BY = b. By chord theorem 6a=2b --> b= 3a. We know that 4(radius)^2 = AB^2 + BC^2 + BX^2 + BY^2 --> 4(50) = 6² + 2² + a² + b² --> 200= 36 +4 + a² + 9a² --> 160 = 10a² --> a=4 AX = 10. Construct OM perpendicular AX. Join OX. MX =AX/2 = 10/2 = 5. In triangle OMX by Pythagoras theorem OM = 5. BM = 1. In triangle OMB by Pythagoras theorem OB = sqrt(26)

We can also use perpendicular chord theorem here. Extend CB and AB to meet the circle at D and E respectively. Now drop perpendiculars OM and ON to chords AE and CD and let the lengths of the perpendiculars be x and y. AM=AB-y=6-y and NC=NB+BC=x+BC=x+2. A perpendicular from the centre to any chord bisects it, so BE=ME-BM=AM-BM=6-2y and BD=NB+ND=NB+NC=2x+2. Using perpendicular chord theorem we get that BD•BC=BA•BE. So 2(2x+2)=6(6-2y). We can now use the radius to get another relation in x and y. In triangle OMA, x^2 + (6-y)^2 = 50. After solving we get x=5 and y=1. OB^2 = x^2 + y^2 = 26. Edit: This is intersecting chord theorem and works for all intersecting chords, not only perpendicular chords.

Ek aur method hai extend BC to cut circle at C' and AB to cut circle at A' let BA'=y then using secant theorem BC' is 3y Now drop perpendiculars to both the chords from centre and use pythagoras and y mil jaayega y mil gya toh agin from secant theorem (OB-r)(OB+r)= y AB hence answer mil gya

Let centre be origin and assume point a be (a,b) and then b will be(a,b-6) and c point will be(a+2, b-6) then the equation of circle will be x^2 + y^2 = 50 and point a and c lies on circle so substitute and solve you will get a value5 and b value 5

Mera method chhota aur easy h, Maine O ko origin let kr liya aur A points ke coordinate likh diya theta variable ke term me √50cos theta,√50sin theta. Phir A coordinate, AB ki length and BC ki help se C point ke coordinate likh diya √50cos theta + 2,√50sin theta-6, ab C point to circle pe lie kr rha h, to C point ke coordinate ko circle ki equation (x²+y²=50) me satisfy krake theta ki 2 values aayi sin theta = 1/√2 or 1/√50. Ab B point ke coordinate usi tarah se likh diya √50cos theta,√50sin theta-6. Ab origin se iski distance hi question me pucha h, wo nikal liya theta ki dono value dalke, ussd OB √26 aur √74 aaya. Ab √74 ho nhi skta kyuki isse B point ki distance center se radius (√50) se bhi jyada ho rhi thi, so the ans is √26. Wese ye likhne me bada h kyuki Maine pura explain kiya h, lekin krne pe bahut Chhota sa h😊

i saw many people assuming the values as constants, pls do it only in mocks or the actual JEE exam, conventional methods take you to a broader understanding of the questions that assuming values can't like cancelling variables on both sides in an inequality.

Literally solved it with one hand while eating maggi Edit: I am in post nut clarity and cringing on my comment rn

My approach was almost same just didnt think abt cosine rule........it really is a nice video and such videos never fail to maintain my courage towards jee advance.......thnks to u i feel adv. Is still doable

Aree mori Maiya 😵‍💫

my approach: join OA,OB,OC. now by pythagoras in triangle ABC we get AB=root(40) and cos(BCA)=1/root(10). Now using this in triangle OAC we can find cos(OCA) by cosine rule. now we can find cos(BCA-OCA)=cos(OCB) by standard identity. Finally simply using cosine rule in triangle OBC we get OB^2=26. I think for the most part soln is similar, just the construction is different

I used a related method as described by others. Extend AB to meet circle at P, join AC. Draw perpendiculars OE and OF on AP and AC. AC = sqrt40 (Pythagoras thoerem), FC = 1/2(sqrt 40). Triangle OFC is similar to Triangular PBC, hence FC/BC=OC/PC, this makes PC = 2sqrt5, and PB = 4 (pythagoras theorem). AP = AB+PB = 10 & EB = 1. From triangle OEP, OE = 5 (Pythogoras theorem). Lastly in triangle OEB, OB = sqrt 26 (Pythagoras theorem). This is with simple geometry (similar triangles, angles of circle & Pythagoras theorem)

My approach probably no one would have thought , Extend BC to cut circle at E and extend AB to cut circle at D now let BD=x and now " Use the property that product of each length of two intersecting chords are equal i.e. AB.BD = CB.BE we get , ED =3x and then draw a perpendicular to chord AD and then use basic properties of circle and equate to opposite sides of reactangle formed between perpendiculars to both chords and enter , it was a cheesey way to the correct ans .

After looking at the problem I realized all I needed is the OAB angle. So at first, I determine the value of angle BAC which is Tan^-1(1/3). Then, angle OAC which is Cos^-1(1/√5) or Tan^-1(2). Then using Tangent formula I determine the angle OAB which is Tan^-1 or 45°. Now using the Cosine formula in triangle OAB I got the answer. OB=√(26).

Draw AC, drop perpendicular from O let @ACB phi tanPhi=6/2=3 and draw line from O to C tan@=rt40/rt10=2. so we get tan(phi-@) as 1/7 so in triangle ocb on using cosine law we have cos(phi-@) two sides and need third side so it will come rt(26).

you are so underrated, this is how jee math should be taught!! especially the fact that you don't really care about "jee" but the beauty of the thing in it❣

OB is a shared side of 2 triangles OBA and OBC, both of whose all *other* sides are known, (OB AB r) and (OB BC r). Also, Angle OBC = 90 + Angle OBA. Let OB = x and theta T = Angle OBA, so Angle OBC becomes 90+T, and so their cosine will be cos(T) and -sin(T) { since cos(90+T) = -sin T } Use cosine rule on all 3 sides to calculate cos(T) and cos(T+90) in terms of constants and OB^2, then use cos^2 T + sin^2 T = 1 to eliminate T and get a quadratic equation of y = x^2 = OB^2 cos T = ( OB^2 + AB^2 - r^2 ) / (2 OB AB) and -sin T = cos(90+T) = ( OB^2 + BC^2 - r^2 ) / (2 OB BC) cos T = ( x^2 + 36 - 50) / 12x = (x/12 - 14 / 12x) and -sin T = (x^2 + 4 - 50) / 4x = (x/4 - 46 / 4x) 1 = cos^2 T + sin^2 T = (x/12 - 7 / 6x)^2 + (x/4 - 23 / 2x)^2 = [ (x - 14/x)^2 + (3x - 138/x)^2 ]/144 => 144 = (1+9) x^2 + (196+19044)/x^2 - (28 + 828 = 856) So, 10 x^4 - 1000 x^2 + 19240 = 0 OR ( x^2 )^2 - 2 (50) x^2 + 1924 = 0, so (x^2 - 50)^2 = 2500 - 1924 = 576 = 24^2, so x^2 = 50 +/- 24 = 74, 26 (OB is smaller than radius, so x^2 < 50) Therefore, *OB^2 = 26*

It can be easily solved using basics of circle like parametric points on circle as a function of any angle

Question direct circles ke property se ban jayega. Extend AB to P and BC to Q to make chords AP and CQ. Now using the theorem for two perpendicular chords in a circle, AB/BC = PB/BQ. Let PB= x then PQ = 3x. Now for perpendicular chords, we have formula 4r² = a²+b²+c²+d², where a,b,c,d are legths of segments of chord. On putting in this formula, we get x=4. So length of AP= 6+4=10. Now draw chord OM perpendicular to AP. Then AM=5. From Pythagoras OM=5. Now join OB to make right traingle OMB. Here MB = AB-AM = 6-5=1 And now apply pythagoras, OB = sgrt(26). Moderate question if someone knows properties of circle well.

It's really interesting but easy than the double disk problem of Mechanics on JEE Advanced 2016 paper.

This question is simple just use intersection chord theorm by extending AB AND CB to meet ag circle

bhaiya maine isko aise kiya:- jo apne diagram banaya voh toh hai hi but maine ek aur prependicular to chord AB dal diya, and foot of prependicular ko E point bol diya and OM and AB ke intersection point ko D BOL DIYA, then angle ACB KO alpha bol diya then angle ADM bhi alpha hoga and angle ODE bhi alpha hoga , then apan ko tanalpha pta hai toh uski madad se apan triangle ADM and triangle ODE mein sin and tan theta lga kr nikallenge

it's been 3 years since i gave my jee advanced, and let me tell you, this question is still a piece of cake. Easiest method is to calculate the angle OCB. to do so, calculate length AC, then calculate angle ACO, also calculate angle ACB, simple trigonometry, now just subtract ACO from ACB, this gives OCB, now use cosine rule of triangle and get the answer, ~sqrt(26.2). No geometric intuition required.

An analogous problem would be- Find the eqn/centre of circle which passes through the intersection of x^2+y^2-2x-6y=0 and its diameter, x/2+y/6=1 and whose radius is √50

Just join o to c oa=oc Calculate ocb angle using basic trigonometry in triangle oac and triangle abc. Obviously you get obc in some form of cos or sine Now since oc, bc and angle ocb known calculate ob using triangle obc.

My approach was as follows :- 1)Extend Line AB till it touches the circle. Take the point at which it intersects the circle as D. 2)Join C and D. Join C and D to the centre of circle. 3)Now, as we see, angle DOC is 2 times of angle DAC. 4)Find angle DAC = BAC (cos(BAC) = AB/AC 5) cos2theta = 2cos^2theta - 1 6)apply cosine rule on cos2theta for triangle DOC. Hence, find CD^2. 7) Using, Pythagoras theorem on triangle DBC to find BD. Add BD and AB to find length of chord AD. 8) Drop a Perpendicular on chord AD. Let the point of intersection of chord and perpendicular be E. 9)By circle's property, we know that AE = 1/2(AD). 10) In triangle OAE, use pythagoras theorem to find length of perpendicular, i.e., OE. 11) find EB = AB-AE. 12) Finally, use pythagoras theorem on triangle OBE to find length of OB.

The distance between A and C is √40 I took O as origin and assumed A to be (√50 ,0) then found the coordinates of the point whose distance from A is √40 by assuming coordinates of that point to be (√50cos theta , √50 sin theta). This is our point C. Upon solving we get the value of theta and therefore coordinates of point C. In triangle ABC, tan of angle ACB is 6/2 = 3 . We know the slope of line AC and using angle between two lines I was able to find the slope of line BC and since we knew coordinates of C I found the equation of line BC. Now using parametric form of line, I found the coordinates of a point whose distance from C along the line BC is 2 units which would give us the coordinates of B as (2√2 , 3√2) Therefore the distance OB was √26 I am comfortable with coordinate geometry so I was able to solve this question quickly ( approx 5 minutes) using coordinate geometry even though it might sound very lengthy.

My approach. Let the coordinate of B be (a,b) so y coordinate a would be (-√(50-a²)) and x coordinate of C would be √(50-b²). So b + √50-a² = 6 and √(50-b²) - a = 2. On solving we get a = 5 and b = 1. So distance = √26

I took C as origin and CA as the x axis. then since we know AB and BC we can get AC with Pythagorean and plot A on the x axis, next we know OA as the radius, so first we draw a perpendicular bisector OM like in the video then plot O right where an arc of length R intersects the perpendicular bisector, we know AM is just half of CA and we also know R so we can get OM with Pythagorean in triangle OAM. we know know the coordinate values for O, the x value is just half of the distance AC and the y value is what we derived just before. Next for B, we will draw two circles around C and A of required lengths AB and BC to get two points of intersection between the circles, the one closer to O is our B, we can get it's co-ordinates by formulating the equation of circles for both the raddi and start points of each of the circles to get then get the system of two equations for two variables and solve them to get our x and y coordinate for B. Now we know the coordinates for both O and B so be can use the distance formula and plug in the co-ordinates directly.

If you use full circle geometry + little vectors you can easily do it Find chord AC aur us diameter vala chota circle banao If done properly we get A(2√10, √10) and AC is lying at an angle to the horizontal (which is connecting Center O and chord AC) Now find coordinates of B by y=y1-6sin(90-A), x=x1-6cos(A) Now as O is considered (0, 0) OB=√(x²+y²) Done paper per karoge tho samaj ayega

I messed up in the first step for taking AB hypotenuse

Mine was an easy approach with only cosine formula Join O to C Let angle(OBA) =x Compute it's cosine in triangle(OAB) Then angle(OBC) = 90+x Compute it's cosine (=-sinx) in triangle OBC Use sin²x + cos²x =1 Solve it you will get OB as √26......

I took O as the origin. And took point A to be the form (√50cos(a), √50sin(a)) where a is positive. And took the point C as (√50cos(b), √50sin(b)).b liesn in the 4th quadrant. Then point B will be (√50cos(a),√50sin(a)-6) and also(√50cos(b)-2, √50 sin(b)). Now we can equate them and we will get 2 eqautions. √50cos(a)=√50cos(b)-2 --------------(i) √50sin(a)-6=√50sin(b) -----------------(ii) U should know how to solve these 2 equations or put it in a calculator or something. Find a or b then wou will get the coordinate then u can find the distance from origin O. My solution is quite simple but involves a bit of calcilation.

Bhaiya, 11-12 minutes mei solve ho gaya, without cosine rule, agar hum perpendicular draw kare from O to AB, I had the same constructions too, usme bas 2-4 places pe Pythgoras lagana padha muje. Loved the question!

Got the qs after hint... creativity at its peak

Bhai kal hi bola tune ki jaldi uthna chahiye aur aj khud hi late kar diya 😢 Vo bhi itna late 😅 This is not acceptable.

Finally you uploaded 😂🎉

Alternate solution: Extend AB to D on the circle. (better complete the circle for clear visualisation) Assume DB to be some x. drop a perpendicular from the center on the chord AD at Y. then first of all find OY by Pythagorean theorem => (3+x/2)^2 + OY^2 = 50. from here we will get OY in terms of x. Also BY=3-x/2. now that we know OY and BY we will get OB=sqrt(50-4x) i.e. in terms of x Now I used a bit of coord geo. WLOG let the center be 0,0. We can write the coordinates of the point C that will be [ -(OY+2), -(BY) ]. We have already calculated OY and BY above in terms of x Now using the equation of circle we will get ---> (OY+2)^2 +(BY)^2 = 50 since the point C lies on the circle and from here on solving we will get x=4. Now OB=sqrt(5-4*6)=sqrt(26) problem I faced --> solving the (OY+2)^2 + (BY)^2 =50 will be a bit of challenge, not because its something very difficult but calculative. I had to use wolfram alpha to reach to x=4. I will be happy to receive suggestions to shorten my solution Thank you!!

Co ordinate geometry se A ko 5,5 assume karke easily ho jayega Aur pure geometry se karne k liye circles k properties pata hone chahiye Isme do properties use honge AB.BD = BC.BE aur a²+b²+c²+d²=4r² where a,b,c,d are the lengths of line segments Bohot dino baad ye property use kar Raha hu Last 9th 10th me use kiya tha

just assume b to be the origin and satisfy c and a in the circle equation with r = sqrt(50) and get circle's center, phir 2 step mei answer aajata jai

I liked the question, thanks for showing that it is solvable this way. Because I tried this way, I wanted the solution to be intuitive but then I gave up and simply went writing the Cartesian form of circle and lines. And then finding the intersecting points, and equating distances with 2 and 6. Then converting point B coordinates to polar form (√50 cos , √50 sin). (Ahh my problem solving skills are rusted)

found the OCB angle, then used simple vector subtraction of CO & CB vector. 🙂

This is old AIME pyq(1983 year) and it's easy for mathematical Olympiad aspirants Here o is centre,Firstly join b and c and then join a and o and finally join o and b Now let angle (bac) = alfa Let angle angle(oab)=t Now drop perpendicular from O to line AC and let d be the foot of perpendicular then let angle (doa) = x then we get angle (oad)=90-x Then we have 90-x= t+alfa , t = 90-(x+alfa) Put cos on bot side to get Cos(t) = sin(x+ alfa) now observe from the given lengths than sin(x) = 1by√5 and sin(alfa) = 1by√10 then we get cos(t) = 1by√2 hence then in the triangle oab using the cosine rule we get here ob²=26 and ob = √26 ( Sorry but this question was not tough)

Thanks bhaiya for questions

Yeh figure toh ek baar me hi soch liya tha lekin fir approach bigad gaya kyoki bohot triangle bana ke pythagorus se kar rha tha, using cosine rule was a good method

can be directly solved by intersection chord theorem..

Joining OB produced to D (on circle). Find sin B using triangle OAB, then OD = BC cos(90-B) = BCsinB. Now, √50-OD=OB. Quadratic aplve.

i did exactly same until the sot part ! but sot nahi ati lol so i was using pythagoras and thoda aur stuff

I took 12 minutes to solve. aprroach similar tha pr maine cosine rule use nhi kiya , normal geometry se angle nikala angle oac aur angle oab fir itf ka use karke angle oab nikal liya. angle oab aaya 45. ab maine perpendicur draw kiya ab pe aur pythagoras lagake uski dist o se nikal li. aur last me firse ek traingle me pythagoras laaya with ob as a hypotenuse. aur bn ans aa gya. itna bhi difficult nhi tha.

Bahit accha question tha aisa hee aur laiye please

took me 20 mins to get it done but I used geometry and half angle subtended at centre by chord property after completing the circle

I am going from 10th to 11th and this took me 1hr. I did it using similarity and Pythagoras theorem.

I did it. It took me 30+minutes because i didn't knew much concepts used in this video, but this question is doable with only similarly of triangles and Pythagoras theorem.

You should have to mention that angle B is a right angle triangle, because it looks like right angle but not mentioned anywhere in question

4 MINUTE SOLUTION: Let the centre of the circle be the origin O(0, 0}, and the coordinates of B be B(x, y). Then the coordinates of A and C are A(x, y+6) and C(x+2, y). Both these points lie on the circle and therefore the coordinates must satisfy the equation of the circle. We thus get 2 equations in 2 unknowns: x^2 +(y+6)^2 = 50 (x+2)^2 + y^2 = 50 Subtracting second from the first, we get x = 3y+8 Putting this back in any of the equations gives a quadratic equation in y: y^2 + 6y + 5 = 0 so y = -1 or -5 so x = 5 or -7 The first solution seems right for the given picture (the second is when C is on the left of O). The coordinates of B are B(5, -1) so OB = sqrt(26).

Try finding OC and put your expression equal to the radius of the circle.(Easiest approach ever) No formula req. 😊

Bhaiya Maine B ko origin man liya Then A aur C ko coordinates se assign Kiya A(0,6) B(2,0) Matlab maine line segment Jo banai thi uske perpendicular bisector per centre of circle tha Usse centre ka co-ordinate a Gaya Then maine centre ke coordinates se origin ka distance nikal liya I mean origin to B hi tha To usse OB ki length a gai

(OB)^2 + (BC)^2 = (radius)^2

That's why I took pencil,eraser,compass,protector and a scale in jee papers Believe me or not 20+ marks are free by constructing the given questions The questions like 1:- find the distance of line from a point 2:- find centre of circle 3:- find the angle between two lines Are easy to solve by constructing and take less than 4 min per question 😅😅😅😂😂

I mean it's not that hard of a question you can also easily solve this by using basic geometry and inverse trigonometry and a little bit of vector addition not much to do, like the data give in such a way that we can find all the angles we need . After solving the the question I also found some unnecessary angles just for fun😂

Bhai agar BC ko extend krke tum uspe o se perpendicular dalo name it (M ) to usko x consider krlo ; ab OM^2+MC^2 = OC^2 karo aur usme MC ko( y+2)maanlo kyunki MB = y consider kiya ab value put krne pe realise hoga ki triangle OMC me pythagoreas theorem aur triangle OMB me pythagoreas theorem compare krne pe Xaur Y ki easiest values 1 aur 5 milegi by hit and trial usse tum triangle OMB me pythagoreas lagao to you waould get OB = 26^.5

Solved using a different approach. Circle ka property hota hai which says two chord let's say AB and CD agar E pe interest kare to AE.BE=CE DE. isko use kiya by extending AB and CB in this case. Fir pythagoras ke alawa aur kisi chiz ki jaroorat nahi pari

The question is easier using cordinate geometry,. Take B as (0,0) C as (2, 0) and A as (0, 6). Not write the equation of diametric circle through AC and+ lamda times equation of line thorugh AC. THen satisfy the condition for radius being rt50. We get two values of lamda for two circles. Now analatycally we reject tha case where we get cordinates of the center of the circle are higher. As the other one is our answer. I gave advance in 2023 and can tell it is not the toughest. The approach he is using is what makes it tough crazy and very out of the box. So knowing this approach is good for later use.

Take O as origin (0,0) A(x,y) B(x,y-6) C(x+2,y-6) OA = radius ----------eq1 OC = radius ----------eq2 x and y me 2 equation Solve karne pe we will get (x,y) = (5,5) Point B (5,-1) OB = √26

This question was also asked in volt test

Itna easy toh tha bhai . My soln : Dono chords ko extend Kiya and used the property PA.PB = PC.PD jo hota hai wo . Fir ek parameter x assume Kiya aur solve Kiya . Ho toh gya . I still say jee adv is tougher

Tbh I did it but it took me 2 hours to figure out but i liked that quality dopamine rush after solving it :)

Thanks

Thought of the approach in under 2 minutes, a long but simple solution is to extend AB to intersect the circle at D, 1. Apply sine rule in triangle ACD, the Circumradius is root50. Figure out length of CD 2. Apply pythagoras on triangle BCD, figure out BD, now you know AD. 3. Construct a perpendicular bisector of AD , call this OE, E lies on AD. 4. Since AD is known, AE is known and hence OE is known, 5. BE is known as well, then find OB, i.e (OE^2 + BE^2)

Solved it using general circle equation by assuming the coordinates for A and C .

Bhaiya koi igdam tigdam calculus ke sawaal solve kijiye na next...waha dikkat aati hai

Bro i gave JEE in 2017, almost forgot many formulas but still i am happy that i am able to solve this question in around 8 mins without any hint 😊

Here the simplest solution AC is radius ,, so AC -BC= root50-2=5.07

Take B as origin,use coord

Calculation tough nahi tha bas construction predict nahi ho rha tha🫤

Simply solved using Pythagoras theorem only! Took two variables x and y and applied pythagoras twice to get two equations of circles and one of the intersection point gave positive x and y and OB² = x²+y², this was how I assumed x and y, this was not even close to advanced it was jee mains level stuff, just complicated it by using cosine rule 😂 (x,y) came out to be (5,1) hence √26 took 2 mins

Bruh I was trying to use length of chord condition on the circle, but then I saw that the chord is incomplete. Also, we can try with similar triangles in a way, by taking the huge triangle and finding the perpendicular distance of O on AB, etc.

by basic algebra and trignometry+cosine rule to u can solve this

If you can't apply basic geometry and construction in some questions it doesn't mean the question is hard compared to JEE. Please look, think and then speak. It took me 1 min to solve the question after looking at it for 1 min.

Hello bhaiya I like your solution but I have a shorter approach, extend both chords to apply power of point theorem (extend AB to X and let BX = p, similarly extend BC to Y in the complete circle now apply power of point so BY = 3p) then we must know that when 4 parts of two intersecting chords intersect at 90 degrees radius of the circle is related to them by 4r^2= a^2+b^2+c^2+d^2 where a b c and d are lengths of those 4 parts and r is radius, then we solve for p it comes to be 4, now drop perpendicular OM on chord AX we know that it would be bisected into two lengths of 5(AM=MX) each then in rt triangle OAM we can find OM to be 5 , lastly in rt triangle OMB, MB=MX-BX= 1 then in rt triangle OMB OB^2= OM^2+ MB^2 => OB = root 26 dekhne me lamba lag rha hain but orally hojata hain solve within short time....

0:45 i am feeling bad for this circle

This is so easy question that it can can be solved by 10th class student. If you will forgot about 11th class coordinate geometry and trigonometry and only think about similarity, area of triangles and appropriate choice of origin and axes then this problem will become very easy.

This was one aime ( usa second level maths Olympiad problem ) problem 3

Bhaiya simple geometry se go raha H done in 7 mins

although I can't explain my solution properly I used inverse trigonometry and the solution of triangles(basic) to solve this question. i constructed OC, i calculated AC using Pythagorean theorem and found the length of perpendicular from O to AC using Pythagorean theorem and then found angle OCA, i also found angle ACB then i used inverse trig to find angle ACB, now using cosine rule i equated angle ACB, OC, BC to find OB.

Can't you just take b as the centre and find the eq of circle passing through (0,6) and (2,0)? isse jo bhi center aayega usse OB nikal jaayega

I'm a NEET aspirant. But I love to take these apparently "impossible" questions asked in JEE advanced as a challenge. I found out the exact solution by using co-ordinate geometry and it was way more simpler than your method.

If my brain explodes due to over info it's his mistake I'll file for a brain insurance

Did it after giving the hint of perpendicular thing

Assuming ABC angle as 90 drawing a circle with A nd C as diametric points P.T. B with centre O' ..Now joined AC and used pythagoras and joined O with AC at O' now calculated OO' and O lies outside the new circle with radius root 10 and centre O' ...Now joined OB and O'B they will be perpendicular using pythagoras in triangle OO'B I'm getting OB as root 30 .... Can someone explain me why I've got the wrong ans ?

My solution was fairly simple, could do it in 10 mins.. Drop a perpendicular from O to AB at say D. Let us say that length of AD is (y) and DB is (6-y). Using Pythagoras in triangle OAD, find the length of perpendicular (h) in terms of (y). Then construct a right angled triangle with vertices O and C passing through D, taking the length of sides as (root50), (6-y), (2+h). In this triangle use Pythagoras to get an equation purely in terms of y (remember, we found h in terms of y). Solve this to get (y=5) and thus (h=5). Lastly, construct right angled triangle ODB and use Pythagoras once again to find the length of the hypotenuse (OB) which comes out to be (root26).