equation of tangent line passing through intersection of two level surfaces

Question

Find the equation of tangent line to the intersection of the two surfaces in $P(4,-2,20)$ $$z=x^2+y^2 \,\,,\,z=2x+4y+20$$

if question like this is given ,i can easily equate both equations ,find gradient and compute the tangent line passing through the point .

But what if surfaces are given as P(x,y,z)=c and Q(x,y,z)=d,with P(x,y,z) and Q(x,y,z) both differentiable how should i find equation of the tangent line passing through a point $(x_0,y_0,z_0)$

Answer 1

Short answer is that a vector parallel to the line is $v=\nabla P \times \nabla Q.$

Since $\nabla P$ is perpendicular to the first surface, $v$ is parallel to it. Since $\nabla Q$ is perpendicular to the second surface, $v$ is parallel to that. So $v$ is parallel to both surfaces, so it must be tangent to the curve.

Answer 2

Let $f(x,y,z)=z-x^2-y^2$. Then $\nabla f(4,-2,20)=(-8,4,1)$ and so the plane tangent to the surface $f(x,y,z)=0$ at $(4,-2,20)$ is$$-8(x-4)+4(y+2)+(z-20)=0\quad\text{or}\quad-8x+4y+z=-20.\tag1$$So, the line that you're after must be contained in the plane $(1)$ and it also must be contained in the plane $z=2x+4y+20$. So, it must be the line$$\{(x,y,z)\in\Bbb R^2\mid-8x+4y+z=-20\wedge z=2x+4y+20\}.$$