Question 272074: Find the equation for the line tangent to the circle x^2+y^2=58 at the points (-3,7)
I dont even know where to begin this problem. thanks for the help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the equation for the line tangent to the circle x^2+y^2=58 at the points
(-3,7)
I dont even know where to begin this problem. thanks for the help.
---------------------------------
The circle is the result of graphing two function:
y = sqrt(58-x^2)
and
y = -sqrt(58-x^2)
====================
The 1st is in the 1st and 2nd quadrants.
The 2nd is in the 3rd and 4th quadrants.
---------------
The point (-3,7) is in the 2nd quadrant so the tangency point
is with y = sqrt(58-x^2)
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Take the derivative of y = sqrt(58 - x^2)
y' = (1/2)(58-x^2)^(-1/2)[-2x]
y'= (-x)/sqrt(58-x^2)
----
Find the slope when x = -3
y'(-3) = 3sqrt(49) = 21
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Find the equation of the line with m = 21 passing thru (-3,7)
7 = 21(-3)+b
b = 70
-----
Tangent Equation:
y = 21x + 70
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Cheers,
Stan H.
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