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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