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gggg

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what are the equation of a line through (8,-3) and passing distance square root of 10 from (3,2)

  • gggg - ,

    To be a "distance of √10 units away from (3,2) means it has to be tangent to the circle
    (x-3)^2 + (y-2)^2 = 10
    or
    x^2 + y^2 -6x - 4y = -3


    let the point of contact be (a,b)
    then
    a^2 + b^2 - 6a - 4b = -3 (#1)

    by slopes...
    slope of radius to (a,b) = (b-2)/(a-3)
    slope from (a,b) to (8,-3) = (b+3)/(a-8)
    but these two lines are perpendicular, so ...
    (b+3)/(a-8) = - (a-3)/(b-2)
    which simplifies to
    a^2 + b^2 - 11a + b = -18 ,(#2)

    subtract #1 - #2
    5a - 5b = 15
    a - b = 3
    a = b+3 (#3)

    sub #3 into #2 to get it simplified to
    b^2 - 2b - 3 = 0
    (b-3)(b+1) = 0
    b = 3 or b = -1
    in #3
    a= 6 or a = 2

    so (a,b) is either (6,3) or (2,-1)

    Now it is easy to find the equation of the two tangents.
    Let me know what you got.

  • gggg - ,

    3x+y-21=0 and x+3y+1

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