The slope of the tangent line is dy/dx.Originally Posted bytnkfub

(x+2)^2 +(y-3)^2 = 29

Differentiate both sides with respect to x,

2(x+2) +2(y-3)(dy/dx) = 0

dy/dx = -(x+2)/(y-3) -----------***

At point (3,5),

dy/dx = -(3+2)/(5-3) = -5/2

Using the point-slope form of the equation of the tangent line,

y-5 = -(5/2)(x-3)

y = -(5/2)(x-3) +5

y = -(5/2)x +15/2 +5

y = -(5/2)x +25/2 ----------answer.

Or,

2y = -5x +25

5x +2y -25 = 0 -------answer.