analytic and solid geometry

a circle is centered at (-3,4) and in tangent to 5x-4y+12=0.find its equation.

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  1. equation must be
    (x+3)^2 + (y-4)^2 = k , were √k is the radius
    2(x+3) + 2(y-4)dy/dx = 0
    dy/dx = -(x+3)/(y-4) which is also the slope of the tangent.
    But from 5x - 4y + 12 = 0 , the slope of the tangent is 5/4

    so (-x-3)/(y-4) = 5/4
    5y - 20 = -4x - 12
    4x + 5y = 8

    4x+5y=8 times 5 ---> 20x + 25y = 40
    5x-4y= -12 times 4 --> 20x - 16y = -48

    subtract: 41y = 82
    y = 2
    then in 4x+5y = 8
    4x + 10 = 8
    x = -1/2 , the point of contact is (-1/2, 2) or (-.5,2)

    so back in original:
    (-1/2 + 3)^2 + (2-4)^2 = k
    k = 41/4 or 10.25

    (x+3)^2 + (y-4)^2 = 41/4

    check:
    at (-.5,2) dy/dx = -(-.5+3)/(2-4) = -2.5/-2 = 5/4

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