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samnatha

  • 3 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

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  1. pansky
    • 3 years ago
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    the equation is \[(x-6)^{2}+(y-2)^{2}=20\]....in order to explain it I need t know whether you have done differentiation?

  2. pansky
    • 3 years ago
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    quick way with differentiation or slightly longer way without...

  3. RadEn
    • 3 years ago
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    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 ??????

  4. samnatha
    • 3 years ago
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    i got it :)

  5. samnatha
    • 3 years ago
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    i got it a completely different way but i got it :)

  6. RadEn
    • 3 years ago
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    good.....but, maybe as a proportion from m then,,,,, take the value of discriminant (D)=0 So, b^2-4ac = 0 (A+2B)^2 – 4(5)C=0 A^2+4AB+4B^2-20C=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 4A-2B+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 = -10B-20 …. (5th eq) Substitute (4th eq) and (5th eq) to (1st eq) gives A^2+4AB+4B^2-20C=0 (3B)^2+4(3B)B+4B^2-20(-10B-20)=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=-10B-20 = -10(-4)-20 = 40-20 = 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^2-12x-4y+20=0

  7. samnatha
    • 3 years ago
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    yupe i got all that thanks for you help :)

  8. pansky
    • 3 years ago
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    @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)

  9. RadEn
    • 3 years ago
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    thanks @pansky... :) yeah, that will work too and easier if we use by differentite, but samnatha not yet learned it

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