Finding a tangent line with implicit differentiatio

Question

Find the tangent line on point $P$ for this curve $(x + 2)^2 + (y - 3)^2 = 37$ on $P(4,4)$

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I tried implicit differentiating

$2(x + 2) + 2(y - 3)y' = 0$

I'm not sure if solving for $y'$ is the way, though

$$y' = \frac{-2(x + 2)}{2(y - 3)}$$ $$y' = -6$$

What do you think?

UPDATE

then

$$y - y_1 = m(x - x_1)$$ $$y - 4 = -6(x - 4)$$ $$y = -6x + 28$$

Answer 1

implicit-differentiation $$f(x,y)=0 \\y'=-\frac{f'_x(x,y)}{f'_y(x,y)}=- \frac{f'_x}{f'_y}$$ if you apply it on $$(x+2)^2+(y−3)^2-37=0=f(x,y)\\y'=-\frac{2(x+2)}{2(y-3)}$$ then put $p(4,4)$ $$y'=-\frac{(x+2)}{y-3}=-\frac{4+2}{4-3}=-6$$