SOLUTION: Find the equation of a line that is tangent to the circle x^2 + y^2 - 8x

SOLUTION: Find the equation of a line that is tangent to the circle x^2 + y^2 - 8x - 8y + 7 = 0 at the point (1,0).

Question 1082354: Find the equation of a line that is tangent to the circle x^2 + y^2 - 8x - 8y + 7 = 0 at the point (1,0).

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the equation of a line that is tangent to the circle x^2 + y^2 - 8x - 8y + 7 = 0 at the point (1,0).
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Take the derivative to find the slope at every point.
2x + 2y*y'-8- 8y
= 0
y' = (8-2x)/(2y-8)
=====
Slope at (1,0)
slope = (8-2)/(0-8) = 6/(-8) = -3/4
---------
Form of line:: y = mx + b
Solve for "b"::
0 = (-3/4)1 + b
b = (3/4)
-----
Equation:: y = (-3/4)x + (3/4)
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Cheers,
Stan H.
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