SOLUTION: Find the equation of the line tangent to the circle x^2+y^2=169 at the point (-5,12).

Question 1041525: Find the equation of the line tangent to the circle x^2+y^2=169 at the point (-5,12).
Found 2 solutions by Alan3354, robertb:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of the line tangent to the circle x^2+y^2=169 at the point (-5,12).
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The slope at any point on a circle centered at the Origin = -x/y
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m = 5/12
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y - 12 = (5/12)*(x+5)

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Note that the point (-5, 12) is on the circle itself.
Differentiating implicitly, we get 2x + 2yy' = 0.
==> x + yy' = 0.
==> -5 + 12y' = 0, or y' = 5/12, the slope of the tangent line.
==> the equation of the tangent line is
<==> 5(x+5) = 12(y-12)
==> 5x + 25 = 12y - 144
==> 5x - 12y + 169 = 0
is the equation of the tangent line at the point (-5, 12).
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