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geometry

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What are the equations of the lines through (-5, -3) and passing at distance 2sqrt5 from (5,7)?

This problem is confusing? How do I solve this? Please I need your Help

  • geometry -

    all points 2√5 from (5,7) form a circle:

    (x-5)^2 + (y-7)^2 = 20

    Now you want a tangent to the circle that passes through (-5,-3). Must be a tangent, because any other line through the circle will come closer than 2√5 to the center.

    So, since lines through (-5,-3) with slope m are

    (y+3) = m(x+5), we need

    (x-5)^2 + (m(x+5)-3-7)^2 = 20
    (x-5)^2 + (mx+(5m-10))^2 = 20
    (m^2+1)x^2 + 10m(m-3)x + (5m-10)^2 - 20 = 0

    Now for the line to be tangent, the above equation must have a single root. That is, the discriminant must be zero:

    100m^2(m-3)^2 - 4(m^2+1)((5m-10)^2-20) = 0
    m = 1/2, 2

    so, the two lines are

    y = 1/2 (x+5) - 3
    y = 2(x+5) - 3

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