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Sab isko itna complicate kyu kar rahe hai X²+Y²=50 (X+2)² + (Y-6)² = 50 Seedha x and y ki value aa jati hai A(5,5) C(7,-1) AND B ka distance pata hai to B(5,-1) ab bas OB nikalna hai √(25+1)=√26
@@aayushchhajed2630 BHAI YEH SAB GALAT HAI AAPKO ITNI INFORMATION HAI HI NAHI KI AAP -2,6 PE CENTRE LE KAR EK CIRCLE BANA DO ROOT50 RADIUS KA AAPNE -2,6 PE CIRCLE KYON BANAYA?
let the cordinates of C be (x, y). Then cordinates of A will be (x-2, y+6). imagine the circle to be centered at origin solve both equations and u will get x, y. its straightforward but messy from there
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
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
Meanwhile I am confused that value of radius is greater than value of "AB" Since, ✓50 > ✓36 than how the diagram shows AB greater than radius of circle. By this conclusion, all the scholars who has taken origin as B(0,0). And considered cente "O" to lie in second quadrant will be false. As then O will lie in third quadrant. Also those who has taken O as origin and considered B to lie in Fourth quadrant is false. As in this it will lie in first quadrant. Hope so, you can understand that diagram is wrong. So prepare diagram yourself and then solve.
me(sees question): arre baap re ye kya hai also me(looks at quesition for 30 seconds): haha ye to laddu sawaal hai answer dekhne se pehle ye meri approach thi: 1. AB ke perpendicular ek radius drop kiya. let point of intersection be M and point on ciircumference be P. 2. OM=OP-BC => OM=√50 -2 3. find AM using pythagoras' theorem. 4. MB= 6-AM 5. find OB using pythagoras' theorem.
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) !!!!
Hello everyone i think i have an easier way to do this question. We need to find the length OB So now join OC Then we have a right angle triangle i.e OBC Now use the Pythagoras theorem. OB sq + BCsq = OC SQ OB sq +4 = 50 OB = root 46 Correct me if i am wrong
I'm glad that I'm through with all this geometry calculus physics shit, and chemistry too. I'm happy with where I am and thinking I've did this back then feels soo amazing about me. Abhi toh 34+45 ke liye bhi calci lagrahi😅
So What I did was Assuming the B as Origin (0,0) C(2,0) And A(0,6) . Now Taking the Center O as (X,Y). Now Using the Distance Formulae I Got. 50= X² + (Y-2)² Now We Know The Y Coordinate Must Should Between 0 and 6 . By Hit and Trial you Got Y can be 1,2,5. And X be 5,6,1. But You Can't take (X,Y) as (6,2). So Either Taking 1 , 5 or 5 , 1. You Get √26. Tadaa 🎉 You Got it By Simple Distance Formula😭
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
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).
This type of problem is actually good for brain excercise but instead of solving this problem i can go with CAD software were i can solve this problem in few seconds
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.
It is totally up to you, how much you can manipulate PEOPLE. That's what this guy did. This problem could easily be done but he chosen the difficult way. Bro you are a JEE Aspirant, I mean you have done coordinate geometry in your syllabus💀 and a circle problem is given to you which even doesn't have any constraints like center coordinates, then why not just let it be on graph with center at origin? (because it could be easily done that way).
Maine hint pane ke baad ek alag method se kiya .. lekin mera answer √50 - 2 aa raha hai .. jiska value 5.079 hai or aapke answer ka numerical value 5.099 , almost same !! Lekin mera 3-4 line mai ho gaya 😂😂
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
Time taken: 8 minutes (Genuinely, see my approach below, jhoot bolne se mera koi fayda nahi hota :D) Method: Pythagoras theorem and properties of circles (Class 9 level math) Not a difficult question, just a bit imaginative.. Bhai aapki videos mujhe bohot acchi lagti hai, but ye starting ka statement galat laga.. 'hona toh nahi hai' bolne se aapne waise hi aadhe logo ka confidence gira diya, and unko genuine try dene se rok diya.. please apni phrasing soch kar kiya karo, many of us look up to you 🙏 My approach: 1) Mark center O, and draw the diameter parallel to the line AB. Let this diameter be PQ, where P is near A and Q is near B. 2) Extend line AB to touch the circle at D. 3) Extend line CB to meet diameter PQ at R. 4) Drop perpendicular from center O to AB at E. 5) Let BD = 2x, BR = y. Thus, OE is also y. 6) Since perpendicular from center bisects the chord, AE = DE = AD/2 = 3+x [This is why I took BD as 2x, not as x, since it would be halved] 7) Also, since ED = 3+x, BD = 2x, thus BE = 3-x, and OR = 3-x 8) In triangle AOE, by pythagoras theorem: OE² + AE² = OA², so, y² + (3+x)² = r² ...........(1) 9) In triangle OBR, by pythagoras theorem: OR² + BR² = OB², so, y² + (3-x)² = OB² ...........(2) 10) In triangle ORC, by pythagoras theorem: OR² + RC² = OC², so, (3-x)² + (y+2)² = r² ...........(3) 11) Now, we have 3 equations, and 3 variables (since r is given), rest is simple algebraic manipulations
Time to solve this, 1 minute. Jyo yeh problem already kiya hain, atleast usko toh saare calculation ke liye toh kam se kam 2 min chahiye. IIT ke sawaal 1st time kisika prob solving approach dekhte nahi. Woh ratta maar maar ke asye hue bacche kar lenge. Magar saccha genuis woh hota jyo isko pehli baar expose kar ke apne dimasg se solution nikaale. Kya karte ye IIT coaching waale? Tonnes of problems do, question papers solve karwaao, ek problem solving pattern dimaag me program karao, aur ussi patterns ka similar cheez istamaal karke JEE crack karo. But in the end agar koi naya pattern wala problem aaya, toh ghanta ek min me solve karenge !
The approach that you took is a complex and tedious approach. This problem can be solved using elementry geometry(Pythagoras theorem) in 2 steps. This is a moderate level PRMO question which we teach to 9th Grade students. Note:- PRMO is the gateway for RMO and then INMO which 9th, 10th and 11th grade students take.
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.
Bro I solved it using coordinate geometry Taking O as (0,0) and taking A as (x,y) then B is (x,y-6) and C is (x+2,y-6) And both and A and C lie on circle so I solved it using general equation It only took me 3 minutes Please reply wether I am correct or wrong
I am a class 12 jee aspirant. Bhaiya aapne jo solution bataya maine nhi dekha kyuki mujhe pata tha mujhe samajh nhi aayega par bhaiya ye question Sirf trigo aur cosine rule ka use karke kiya maine aur ans match hone ki khushi to kya hi batau. Gave 1 hr to the question
Bhai B right angle hey bola hi nahi apney wahi hogai confusion figure mey to dikhi nahi perpendicular sign or itni choti figure mey perpendicular kese manley 😅
Well 😢 I thaught about this idea to solve but 😂 u know this was first I didn't knew chord rule , didn't knew cosine rule 😊 😂 So I stopped and watch video in 4x speed 😂 Now I am happy 😊 Thanks brother ❤
itne aasan sawal ko complex karke koi bhi difficulty badha sakta hai... it was an easy question, aisa nahi ki hoga hi nahi solve. Pythagoras se ho jata hai honestly. Find better questions thank you
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!!
From pure geometry, Extend AB to meet the circle at X and BC to meet the circle at Y. BY/BX = AB/BC = 3 from similarity Let BX = x and BY = 3x Now draw perpendicular from O to BY, name M and O to AX name N. MY = (BY + BC)/2 = (3x + 2)/2 ON = MB = BY - MY = (3x - 2)/2 AN = (AB + BX) / 2 = (6 + x)/2 ON² + AN² = OA² = r² This will give the value of x as 4. Then, NB = AB - AN = 6 - (6 + 4)/2 = 1 ON = (3 × 4 - 2)/2 = 5 OB² = ON² + NB² OB = √26 The solution seems long but it takes less time to process each step.
You can solve purely geometrically. Choose origin at O = (0, 0) Assume without loss of generality that AB is vertical (parallel to Y axis), which means that BC is horizontal (parallel to X axis) A = (a, b) B = (a, b-6) C = (a+2, b-6) Equation of circle is x^2 + y^2 = 50 A is on the circle, which means that a^2 + b^2 = 50 C is on the circle, which means that (a+2)^2 + (b-6)^2 = 50 a^2 + 4a + 4 + b^2 - 12b + 36 = 50 (a^2 + b^2) + 4a - 12b + 40 = 50 (50) + 4a - 12b + 40 = 50 a - 3b + 10 = 0 a = 3b - 10 (3b - 10)^2 + b^2 = 50 9b^2 - 60b +100 + b^2 - 50 = 0 10b^2 - 60b + 50 = 0 b^2 - 6b + 5 = 0 (b-5)(b-1) = 0 b = 5 or b = 1 (a, b) = { (5, 5), (-7, 1) } OB = sqrt((a-0)^2 + (b-6-0)^2) OB^2 = 5^2 + (-1)^2 OB^2 = 26 OB = sqrt(26) *or* OB = sqrt((a-0)^2 + (b-6-0)^2) OB^2 = (-7)^2 + (-5)^2 OB^2 = 74 OB = sqrt(74) OB is required to be less than r = sqrt(50), since B is a point inside the circle. Hence this is an extraneous solution. Therefore the answer is just *OB = sqrt(26)*
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
To all those people happy after solving the problem JEE Advanced problems are much tougher than these Ye Halwa hai It can be called a JEE Mains level problem though
Hmm, Its a pretty subjective opinion. But yes, many students cracked it through approaches way easier than I did. in that way, I’d say mains level jaa sakta.
Bro what are you saying I have practically just finished 10th and some how I know cosine rule and cos(A±B) expansion it's the only thing required to solve this question once you told angle ABC WAS 90 Degrees
Bhai maine bhi ise solve kiya By using cooridinate geometry O-》(0,0) A-》(X1,Y1) B-》(X1,Y2)(BY GRAPH I plug x1) C-》(X2,Y2)(BY Graph) And just created eq. By the given data And try to solve I get x1=5,y2=-3 And apply distance formula But I get sqrt(34) as the answer But I know my method is easy and correct but what went wrong no body know please help me bro🎉🎉🎉🎉🎉🎉🎉
