Proof that any tangent plane to $z=-\sqrt {1-x^2-y^2} $is normal at point $P$ to $\vec{OP}$ Proof that the tangent plane of f(x,y) in a point $P_0(x_0,y_0,f(x_0,y_0))$ to a vector $\vec {OP} $ given
$$f(x,y)=-\sqrt {1-x^2-y^2}$$
First the domain is defined for
$x^2*y^2\lt 1$
Now I find the partial derivatives
$$\frac {\partial f}{\partial x}=\frac {x}{\sqrt{1-x^2-y^2}}$$
$$\frac {\partial f}{\partial y}=\frac {y}{\sqrt{1-x^2-y^2}}$$
The tangent plane to the point $P_0$ is
$$z+\sqrt{1-x_0^2-y_0^2}=\frac {x_0}{\sqrt{1-x_0^2-y_0^2}}(x-x_0)+\frac {y_0}{\sqrt{1-x_0^2-y_0^2}}(y-y_0)$$
$$0=\frac {1}{\sqrt{1-x_0^2-y_0^2}}(x_0(x-x_0)+y_0(y-y_0))-( z+\sqrt{1-x_0^2-y_0^2})$$
I know that in order to $\vec {OP}$ to be normal to the plane $\vec{OP}\bullet \vec {V}=0$ where $\vec {V}$
 is any vector contained in the tangent plane, I seem to be struggling to find this vector and finish the proof.
 A: For a geometric perspective, observe that the equation $f(x,y)=-\sqrt {1-x^2-y^2}$ can be rewritten as $x^2 + y^2 +z^2 =1$ which is the equation of the unit sphere centered at origin. It should be clear from geometry that any tangent plane to the sphere must be normal to the radius at the point of contact.
If vectorial methods are desired, note that if we rewrite the function as $x^2 + y^2 +z^2 -1 = 0$, then the gradient of $w = x^2 + y^2 +z^2 -1$ must be perpendicular to any 3d curve given by $x^2 + y^2 +z^2 -1 = k$ for some constant $k$. Now the gradient is $\left[ 2x,2y,2z \right] = 2\left[ x,y,z \right] = 2\vec{OP}$. Thus any tangent plane to $x^2 + y^2 +z^2 -1 = 0$ ( and hence $f(x,y)=-\sqrt {1-x^2-y^2}$) is normal to the gradient of $w$ and hence the radius vector $\vec{OP}$.
A: $\vec {OP}=\left(x_0,y_0,-\sqrt{1-x_0^2-y_0^2}\right)$
The vector normal to the tangent plane is $\vec n$
$\vec {n}=\left(\dfrac{x_0}{\sqrt{1-x_0^2-y_0^2}},\dfrac{y_0}{\sqrt{1-x_0^2-y_0^2}},-1\right)$
$\left(\vec i,\;\vec j,\;\vec k\right)$ is the base
and cross product is
$$\vec{OP}\times \vec n=\det\left| 
\begin{array}{ccc}
 \vec i & \vec j & \vec k \\
 a & b & -\sqrt{-a^2-b^2+1} \\
 \dfrac{a}{\sqrt{-a^2-b^2+1}} & \dfrac{b}{\sqrt{-a^2-b^2+1}} & -1 \\
\end{array}
\right|=\vec O$$
thus $\vec{OP}\parallel \vec n$ which means that $\vec{OP}$ is normal to the tangent plane.
Hope this helps
