SOLUTION: Find the equation of the lines which are tangent to the circle x^2+y^2=625 & perpendicular to the line 24x-7y+84=0

Question 872883: Find the equation of the lines which are tangent to the circle x^2+y^2=625 & perpendicular to the line 24x-7y+84=0
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of the lines which are tangent to the circle x^2+y^2=625 & perpendicular to the line 24x-7y+84=0
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m, the slope of 24x-7y+84=0 is 24/7
The slope perpendicular = -7/24
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The slope at any point on a circle about the Origin is -x/y
-7/24 = -x/y
7y = 24x --> y = 24x/7
x^2+y^2=625
x^2 + (576x^2/49) = 625
49x^2 + 576x^2 = 30625
625x^2 = 30625
x^2 = 49
x = 7, y = 24
x = -7, y = -24
Use the points and the slope (-7/24) to find the eqns of the 2 lines.
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