Find the tangent lines to the given circle as directed. x^2+y^2=34 at (-3,5).
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calendarEducator since 2008
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Given the equation of the line :
x^2 + y^2 = 34
We need to find the tangent line at the point ( -3,5)
First we will differentiate with respect to x.
==> 2x + 2yy' = 0
==> 2yy' = -2x
==> y' = -2x/2y
==> y' = -x/y
Now we will substitute with the point (-3,5) to find the slope.
==> m = y' = -(-3)/5 = 3/5
Then the equation of the line is given by :
y-y1 = m(x-x1)
==> y-5 = (3/5) (x+3)
==> y= (3/5)x + 9/5 + 5
==> y= (3/5)x + 34/5
Then the equation of the line is:
==> 5y - 3x -34 = 0
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calendarEducator since 2010
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The slope of the tangent to a curve f(x) is the value of the first derivative of the curve for the appropriate x.
Here we have x^2 + y^2 = 34
Differentiating both the sides 2x + 2y (dy/dx) = 0
=> dy/dx = -2x/2y = -x/y
At x = -3 and y = 5
dy/dx = 3/5
The equation of the tangent is (y - 5) / (x + 3) = 3/5
=> 5y - 25 = 3x + 9
=> 3x - 5y + 34 = 0
The required equation of the tangent is 3x - 5y + 34 = 0