I thought this was a lil to simple and think my answer might be wrong :/

Determine the equation of the tangent to the curve at the given point

(x+2)^2+(y-3)^2=29 P(3,5)

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- Apr 5th 2006, 01:17 PMtnkfubFind the equation,
I thought this was a lil to simple and think my answer might be wrong :/

Determine the equation of the tangent to the curve at the given point

(x+2)^2+(y-3)^2=29 P(3,5) - Apr 7th 2006, 02:34 PMticbolQuote:

Originally Posted by**tnkfub**

(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.