Tangent line to a two variables function

Question

Given the function $$f(x,y)=\sqrt{x^2+y^2-9}$$ I have to find the tangent line to it obtained from the intersection of the plane $y=-3$ with its graph at $(4,-3,4)$.

Since $$\frac{\partial f}{\partial x}(4,-3)=1$$ then the tangent line will be $$y-(-3)=1\cdot(x-4)$$

I want to know if it's correct.

Answer 1

The intersection curve of the surface given by $f(x,y) = \sqrt{x^2+y^2-9}~ $ and plane $y = -3$ is in fact a pair of lines.

Plugging in $y = -3$, we get $z = f(x,y) = |x|$
(note $f(x,y) \geq 0)$

And point $(4, -3, 4)$ is on line $z = x$. So the equation of tangent line is $z = x, y = -3$

Answer 2

We have that at $y=-3$

$$z=f(x,-3)=\sqrt{x^2+y^2-9}=\sqrt{x^2}=|x|$$

and the tangent line, in the $z-x$ plane, should be

$$z-4=1\cdot (x-4) \implies z=x$$