find the equation of tangent line of (x^2)+(y^2)+(2x)-y-17=0 at (3,1)
thanks.

Question

anonymous · Answer

U have centre of circle and (3,1) so get that line there and other is perpendicular through (3,1)

anonymous · Answer

thank you.

anonymous · Answer

use the chain rule to get the derivative.

$$\frac{d}{dx}(x^2+y^2+2x-y-17) = 2x+2y\frac{dy}{dx}+2-\frac{dy}{dx}$$
$$= \frac{dy}{dx}(2y-1)+2x+2 = 0 \iff \frac{dy}{dx} = -\frac{2x+2}{2y-1}$$
plug in the point (3,1) to get your slope.  then use the point-slope formula to get your line.

anonymous · Answer

i can't understand!

anonymous · Answer

U know the circle centre, right?

anonymous · Answer

...no I can't understand the codes... because earlier the euqtion you had econded didn't load properly

anonymous · Answer

U get centre by completing square so u can see...

(x+1)^2 and (y-1/2)^2.....-> (-1,1/2) for the centre.

Use that with (3,1) to get slope, then the perpendicular through (3,1) has negative reciprocal of that slope.

(I am assuming u know formulas for slope and equation of line).