SOLUTION: how do i find the equation of the line tangent to the circle x^2+y^3=13 at the point (-2,3). write the answer in slope-intercept form.
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Question 665773: how do i find the equation of the line tangent to the circle x^2+y^3=13 at the point (-2,3). write the answer in slope-intercept form. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! x^2+y^3=13
The co ordinates of center are (0,0)
point on circle = (-2,3)
The slope of radius = (0-3)/(0-(-2))
= -3/2
The tangent is always perpendicular to the radius.
so slope of tangent will be 2/3 ( negative reciprocal)
the tangent passes through (-2,3)
m= 2/3
Y = m x + b
3.00 = 2/3 * -2 + b
3.00 = -1 1/3 + b
b= 3 + 1 1/3
b= 13/ 3
So the equation will be
Y = 2/3 x + 13/3
m.ananth@hotmail.ca
