SOLUTION: Find the equation of the line tangent to the circle at the indicated point. x^2+y^2-6x+8y-144 = 0 at point (8,8)

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Question 1179955: Find the equation of the line tangent to the circle at the indicated point.
x^2+y^2-6x+8y-144 = 0 at point (8,8)
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!










centre is (,) and the radius is
slope of line through (,) and (,)



since tangent is perpendicular to it, the slope of line of the tangent is therefore
So line is
So, tangecy point is at (,)


=> your tangent



Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.

You can directly differentiate the equation


    2x*dx + 2y*dy -6*dx + 8*dy = 0

    (2x-6)dx + (2y+8)dy = 0

    (2x-6)dx = -(2y+8)dy

     = 


Now substitute the given values (coordinates) into the formula to get the slope


     =  =  = -.


So, now you know the point (8,8) and the slope .


Hence, an equation of the tangent line is


    y - 8 = .      ANSWER


And you can transform it equivalently to any other form you wish.

Solved.



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