Find an equation of the tangent line to the curve of intersection of the surface $z=f(x,y)=16-x^{2}-y^{2}$:

Solution  (a) The slope of the tangent line to the curve of intersection of $z=16-x^{2}-y^{2}$ and the plane $y=2$ at any point is $f_{x}(x,2)=-2x$. At the point $(1,2,11)$, the slope is $f_{x}( 1,2) =-2(1) =$ $-2$. This tangent line lies on the plane $y=2$. Symmetric equations of the tangent line are $$z-11= \frac{x-1}{-1/2}\qquad y=2$$

(b) The slope of the tangent line to the curve of intersection of $z=16-x^{2}-y^{2}$ and the plane $x=1$ at any point is $f_{y}(1,y)=-2y$. At the point $(1,2,11)$, the slope is $f_{y}( 1,2) =-2( 2) =-4$. This tangent line lies on the plane $x=1$. Symmetric equations of the tangent line are $$z-11= \frac{y-2}{-1/4}\qquad x=1$$