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Math Help - equation of a circle!

  1. #1
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    equation of a circle!

    ok so i got the tangent lines of the two points,

    point A(4,3) y=4/3x-25/3

    point B(4,-3) y=-4/3x+25/3

    now i just need help finding the normal lines...thanks...

    mathlete

    The normal line to a curve at a point is perpendicular to the tangent line at that point.
    Find the equation of the normal line to this circle at Point A.
    y(x) =
    Find the equation of the normal line to this circle at Point B.
    y(x) =

    (d) At what point do the two normal lines intersect?
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  2. #2
    Super Member wingless's Avatar
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    Quote Originally Posted by mathlete View Post
    ok so i got the tangent lines of the two points,
    point A(4,3) y=4/3x-25/3

    point B(4,-3) y=-4/3x+25/3
    Are they
    A(4,3)~, ~y=\frac{25}{3} - \frac{4x}{3}
    and
    A(4,-3)~, ~y=\frac{4x}{3}-\frac{25}{3}
    ?

    First,
    A(4,3)~
    ~y=\frac{25}{3} - \frac{4x}{3}
    The line has the slope -\frac{4}{3}.
    So the line perpendicular to it should have the slope \frac{3}{4}

    The general formula is,
    y = mx + C, m is the slope, C is a constant
    y = \frac{3}{4}x + C
    It must pass A(4,3)
    3 = \frac{3}{4}4 + C
    C = 0
    So the equation of the normal is,
    y = \frac{3}{4}x

    You can apply the same for the other line.
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