Let $f(x,y) = -(1-x^2-y^2)^{1/2}$ for $(x,y)$ such that $x^2+y^2 < 1$. Show that the plane tangent to the graph of $f$ at $(x_0,y_0,f(x_0,y_0))$ is orthogonal to the vector $(x_0,y_0,f(x_0,y_0))$.
I am confused about the overall steps of solving such a problem. The tangent plane is given by $z = f(x_0,y_0) - \frac {x_0} { \sqrt {1-x_0^2-y^2}}(x-x_0) - \frac {y_0}{ \sqrt {1-x_0^2-y_0^2}}(y-y_0)$, which has normal $( \frac {x_0} { \sqrt {1-x_0^2-y^2}} , \frac {y_0} { \sqrt {1-x_0^2-y^2}}, -1)$, and here I am stuck. Perhaps I have gone astray, any hints appreciated.