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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
 one year ago
 one year 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
 one year ago
 one year ago

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

panskyBest ResponseYou've already chosen the best response.0
quick way with differentiation or slightly longer way without...
 one year ago

RadEnBest ResponseYou've already chosen the best response.1
well, @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 ??????
 one year ago

samnathaBest ResponseYou've already chosen the best response.0
i got it a completely different way but i got it :)
 one year ago

RadEnBest ResponseYou've already chosen the best response.1
good.....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
 one year ago

samnathaBest ResponseYou've already chosen the best response.0
yupe i got all that thanks for you help :)
 one year ago

panskyBest 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)
 one year ago

RadEnBest ResponseYou've already chosen the best response.1
thanks @pansky... :) yeah, that will work too and easier if we use by differentite, but samnatha not yet learned it
 one year ago
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