Tangent Plane to sphere using 3D

Question

How to find the tangent plane to sphere say, $x^{2} + y^{2} + z^{2} +2ux+2vy+2wz+d=0$ from the point $P(x1,y1,z1)$ not necessarily on the sphere.

If possible, solve using 3D not vectors.

Answer 1

Hint: You can calculate a second point, which would satisfy the equation of a plane. Find the second point by using the vectors: $\mathbf{x}',\mathbf{y'}, \mathbf{z'}$. $$\mathbf{x}'=\dfrac{\mathbf{x}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}, \mathbf{y}'=\dfrac{\mathbf{y}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}, \mathbf{z}'=\dfrac{\mathbf{z}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}$$

Each of this vectors should satisfy the following equation:

$$(x_1+u)^2+(y_1+v)^2+(z_1+w)^2-[\mathbf{x'}(x_1+u)+\mathbf{y'}(y_1+v)+\mathbf{z'}(z_1+w)]^2=r^2$$

Applying this to the sphere $x^{2} + y^{2} + z^{2} +2ux+2vy+2wz+d=0$ gives: $$x_1^2+y_1^2+z_1^2+2x_1u+2y_1v+2z_1w+d-[\mathbf{x'}(x_1+u)+\mathbf{y'}(y_1+v)+\mathbf{z'}(z_1+w)]^2=0$$