math question

Discussion in 'Community Discussion' started by wonga1127, Aug 30, 2007.

  1. wonga1127 macrumors 6502

    Joined:
    Mar 16, 2006
    Location:
    Wishing for a magic bus.
    #1
    Okay, having trouble with this problem.

    Show that the equation of the line tangent to the circle x^2 + y^2 = a^2 at the point (X1 , Y1) (thats x sub 1 y sub 1) on the circle is X1(x) + Y1(y) = a^2

    Where do I even start?
     
  2. zap2 macrumors 604

    zap2

    Joined:
    Mar 8, 2005
    Location:
    Washington D.C
    #2
    Oh, my. I should know how to do this...



    but I've forgotten, sorry.

    Someone here is bound to be able to help
     
  3. siurpeeman macrumors 603

    siurpeeman

    Joined:
    Dec 2, 2006
    Location:
    the OC
    #3
    the slope from the center (0,0) to the point of tangency (x1,y1) is equal to y1/x1. the line from the center to the point of tangency is perpendicular to the tangent line, so the slope of the tangent line is equal to the negative reciprocal, or -x1/y1. using point-slope formula, you get an equation of y - y1 = (-x1/y1)(x - x1).

    multiplying and rearranging will lead you to (x1)*x + (y1)*y = (x1)^2 + (y1)^2. looking at the original equation, (x1)^2 + (y1)^2 is equal to a^2, so substituting will get you your equation of (x1)^2 + (y1)^2 = a^2.

    you're welcome. ;)
     
  4. yg17 macrumors G5

    yg17

    Joined:
    Aug 1, 2004
    Location:
    St. Louis, MO
    #4
    [​IMG]
     
  5. wonga1127 thread starter macrumors 6502

    Joined:
    Mar 16, 2006
    Location:
    Wishing for a magic bus.
    #5
    thank you.

    And the lolcat made me laugh.

    edit: okay i did the point slop formula and everything, but when I simplfied and rearranged I got y1y + x1x = x1^2 + y1^2. x^2 +y^2 = a^2. im stuck!
     

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