anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the equation is \[(x-6)^{2}+(y-2)^{2}=20\]....in order to explain it I need t know whether you have done differentiation?
anonymous
  • anonymous
quick way with differentiation or slightly longer way without...
RadEn
  • RadEn
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 ??????

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i got it :)
anonymous
  • anonymous
i got it a completely different way but i got it :)
RadEn
  • RadEn
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
anonymous
  • anonymous
yupe i got all that thanks for you help :)
anonymous
  • anonymous
@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
  • RadEn
thanks @pansky... :) yeah, that will work too and easier if we use by differentite, but samnatha not yet learned it

Looking for something else?

Not the answer you are looking for? Search for more explanations.