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)

Algebra ->  Average -> 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)      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E2-6x%2By%5E2%2B8y=144+
%28x%5E2-6x%2Bb%5E2%29-b%5E2%2B%28y%5E2%2B8y%2Bb%5E2%29-b%5E2=144+
%28x%5E2-6x%2B3%5E2%29-3%5E2%2B%28y%5E2%2B8y%2B4%5E2%29-4%5E2=144+
%28x-3%29%5E2-9%2B%28y%2B4%29%5E2-16=144+
%28x-3%29%5E2%2B%28y%2B4%29%5E2-25=144+
%28x-3%29%5E2%2B%28y%2B4%29%5E2=144+%2B25
%28x-3%29%5E2%2B%28y%2B4%29%5E2=169
%28x-3%29%5E2%2B%28y%2B4%29%5E2=13%5E2
centre is (3,-4) and the radius is r=13
slope of line through (3,-4) and (8,8)
m=%288-%28-4%29%29%2F%288-3%29
m=%288%2B4%29%2F5
m=12%2F5
since tangent is perpendicular to it, the slope of line of the tangent is therefore -1%2F%2812%2F5%29=-5%2F12
So line is y=-%285%2F12%29x%2Bb
So, tangecy point is at (8,8)
8=-%285%2F12%298%2Bb
b+=+34%2F3
y=-%285%2F12%29x%2B34%2F3=> your tangent



Answer by ikleyn(52787) About Me  (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

    %28dy%29%2F%28dx%29 = -%282x-6%29%2F%282y%2B8%29


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


    %28dy%29%2F%28dx%29 = -%282%2A8-6%29%2F%282%2A8%2B8%29 = -10%2F24 = -5%2F12.


So, now you know the point (8,8) and the slope -5%2F12.


Hence, an equation of the tangent line is


    y - 8 = %28-5%2F12%29%28x-8%29.      ANSWER


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

Solved.