# 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.

Algebra ->  -> 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.      Log On

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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(13127)   (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