A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
please help
A circle hast he line y = 2x as a tangent at the point (2, 4 ). The circle circle also passes through the point (4 , 2).
find the equation of the circle
anonymous
 4 years ago
please help A circle hast he line y = 2x as a tangent at the point (2, 4 ). The circle circle also passes through the point (4 , 2). find the equation of the circle

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the equation is \[(x6)^{2}+(y2)^{2}=20\]....in order to explain it I need t know whether you have done differentiation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0quick way with differentiation or slightly longer way without...

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1well, @samnatha.... i hope u can understand what i write !!! u can asking to me which parts u didnt understanding it.... The equation of circle : X^2+y^2+Ax+By+C=0 and Tangent line of circle given that y=2x Now, joining the 2 curves gives X^2+(2x)^2+Ax+B(2x)+C=0 X^2+4x^2+Ax+2Bx+C=0 5X^2+(A+2B)x+C=0 so far so good ??????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got it a completely different way but i got it :)

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1good.....but, maybe as a proportion from m then,,,,, take the value of discriminant (D)=0 So, b^24ac = 0 (A+2B)^2 – 4(5)C=0 A^2+4AB+4B^220C=0 …. (1st eq) Back to equation of circle : x^2+y^2+Ax+By+C=0 Passes through (2,4), it means (2)^2+(4)^2+A(2)+B(4)+C=0 or 2A+4B+C+20=0 …. (2nd eq) Passes through (2,4), it means x^2+y^2+Ax+By+C=0 (4)^2+(2)^2+A(4)+B(2)+C=0 or 4A2B+C+20=0 …. (3rd eq) If (2nd eq) – (3rd eq) gives 2A+6B=0 2A=6B A=3B …. (4th eq) Substitute (4th eq) to (2nd eq) gives 2A+4B+C+20=0 2(3B)+4B+C+20=0 10B+C+20=0 C = 10B20 …. (5th eq) Substitute (4th eq) and (5th eq) to (1st eq) gives A^2+4AB+4B^220C=0 (3B)^2+4(3B)B+4B^220(10B20)=0 9B^2+12B^2+4B^2+200B+400=0 25B^2+200B+400=0 (divided by 25 to both sides) B^2+8B+16=0 (B+4)(B+4)=0 So, get B =  4 Substitute B =  4 to (4th eq) and (5th eq) A = 3B = 3(4) = 12 C=10B20 = 10(4)20 = 4020 = 20 And as finishing, subtitute the value of A,B, and C to the equation of circle : x^2+y^2+Ax+By+C=0 Thus, x^2+y^212x4y+20=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yupe i got all that thanks for you help :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@RadEn nice algebra!! you could have used the fact that the perpendicular bisector of a chord goes through the centre of the circle and find it from chord between (2,4) and (4,2)

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1thanks @pansky... :) yeah, that will work too and easier if we use by differentite, but samnatha not yet learned it
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.