SOLUTION: Find all points on the circle x^2 + y^2 = 81 where the slope is 9/40. (x, y) = ( , ) (positive y-coordinate) (x, y) = ( , ) (negative y-coordinate) I've been trying everyt

Question 1157854: Find all points on the circle x^2 + y^2 = 81 where the slope is 9/40.
(x, y) = ( , ) (positive y-coordinate)
(x, y) = ( , ) (negative y-coordinate)
I've been trying everything and attempting to solve based on similar questions, please include all steps so I also know how to solve afterwards
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

That means:

Find all points on the circle x^2 + y^2 = 81 where the slope OF A LINE DRAWN
TANGENT TO THE CIRCLE AT THAT POINT is 9/40.





We take the derivative implicitly



Divide every term by 2



Substitute 9/40 for (dy)/(dx)



Solve for y:



Multiply both sides by 40/9



Substitute for y in the equation of the circle:





Multiply through by 81









Substituting for x² in the equation of the circle:





Multiply through by 1681















So the two points where the slope of the tangent line is 9/40 are:



and



Edwin

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