Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$ I have to solve the following statements, finding the tangent planes to $f$ and its points of tangency:

that contains the line in $\mathbb{R}^3$ that passes through the points $(3,0,3)$ and $(0,-3,3)$

and this other statement:

that contains the point $(0,0,2)$

For the first statement, I tried first making the parametric equation and I found that $L(t) = (3-3t, -3t, 3)$, and I also know that the equation of the plane is given by $$p(x,y) = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) = -2x_0 (x-x_0) -2y_0 (y-y_0) + (1 -x_0^2 - y_0^2)$$ but after that, I don´t see what else to do. Could you give me any hint for both statements?
 A: The partial derivatives are:$$f'_x = -2x, \quad f'_y = -2y$$
Thus, we have
$$\begin{align}z &= \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) \\ 
&= -2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1\tag{1}\end{align}$$
Since $z(0,0) = 2$, we have
$$x_0^2 +y_0^2 + 1 = 2 \implies x^2_0 + y_0^2 = 1 \tag{2}$$
From this, $(1)$ simplifies to
$$\begin{align}z&=-2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1 \\&= -2x_0x-2y_0y+x_0^2+y_0^2+1 \\&= -2x_0x-2y_0y + 2 \tag{3}\end{align} $$

For all points $(x_0,y_0)$ satisfying $(2)$, the equation $(3)$ gives a tangent plane that passes through $(0,0,2)$: For example, when $x_0 = \sqrt{\frac{2}{3}}$ and $y_0 = \sqrt{\frac{1}{3}}$ you get the following picture:


One more example with another random choice of a tangency point satisfying $(2)$:

A: Given
$$
\cases{
f(x,y,z)=x^2+y^2+z-1=0\\
\pi(x,y,z) = a x+b y + z c + d=0\\
p=(x,y,z)\\
p_1=(3,0,3)\\
p_2=(0,-3, 3)
}
$$
Construct $\pi(x,y,z)$ such that it contains the line $L \to p = p_1+\mu(p_2-p_1)$  and is tangent to $f(x,y,z)$.
At the tangency point $p_0$ we have
$$
\cases{
\nabla f(p_0) = \lambda \nabla \pi(x_0)\\
f(p_0) = 0\\
\pi(p_0) = 0\\
\pi(p_1) = 0\\
(p_2-p_1)\cdot\nabla \pi(p_0) = 0\\
\|\nabla \pi(p_0)\|=1
}
$$
Here
$$
\cases{
\nabla f(p_0) = (2x_0,2y_0,1)\\
\nabla\pi(p_0) = (a,b,c)
}
$$
so we have $8$ equations on the $8$ unknowns $x_0,y_0,z_0,a,b,c,d,\lambda$. Solving those equations we have two solutions:
$$
p_0 = \left\{
\begin{array}{ccc}
 (\frac{1}{2} \left(3-\sqrt{13}\right), & \frac{1}{2} \left(\sqrt{13}-3\right), & 3 \sqrt{13}-10) \\
( \frac{1}{2} \left(3+\sqrt{13}\right), & -\frac{1}{2} \left(3+\sqrt{13}\right), & -10-3 \sqrt{13})\\
\end{array}
\right.
$$
and the planes are characterized by
$$
(a,b,c,d) = \left\{
\begin{array}{cccc}
 (-\sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \frac{1}{68} \left(\sqrt{102}
   \left(9-\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9-\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9-\sqrt{13}\right)}-\sqrt{102}
   \left(9-\sqrt{13}\right)^{3/2}\right)) \\
 (-\sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)} ,& \frac{1}{68} \left(\sqrt{102}
   \left(9+\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9+\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9+\sqrt{13}\right)}-\sqrt{102}
   \left(9+\sqrt{13}\right)^{3/2}\right)) \\
\end{array}
\right.
$$
Attached a plot showing in red the given line and in black the tangentcy points.

