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Thread: 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
    $\displaystyle A(4,3)~$, $\displaystyle ~y=\frac{25}{3} - \frac{4x}{3}$
    and
    $\displaystyle A(4,-3)~$, $\displaystyle ~y=\frac{4x}{3}-\frac{25}{3} $
    ?

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

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

    You can apply the same for the other line.
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