Take two families of circles $$ (x-c_1)^2+y^2=c_1^2\qquad\mbox{and}\qquad x^2+(y-c_2)^2=c_2^2\;, $$ where $c_1$ and $c_2$ are positive constants. Use the gradient to show that these two families intersect orthogonally.
Attempt: Let $f(x,y)=(x-c_1)^2+y^2-c_1^2$ and $g(x,y)=x^2+(y-c_2)^2-c_2^2$. Then $\nabla f (x,y)=(2(x-c_1),2y)$ and $\nabla g(x,y)=(2x,2(y-c_2))$. However, I'm not sure how one could go about establishing that $$ \nabla f(x,y)\cdot\nabla g(x,y) = 0. $$ I have thought about working in polar coordinates, but this has not helped either. Any ideas, please?
First establish some useful relations at intersection points, per @SecretMath's comment. Namely $$ \left \{ \begin{array}{ccc} (x-c_1)^2 + y^2 & = & c_1^2 \\ x^2 + (y-c_2)^2 & = & c_2^2 \end{array} \right . \implies \left \{ \begin{array}{ccc} x^2 - 2c_1x + y^2 & = & 0 \\ x^2 + y^2 - 2c_2y & = & 0 \end{array} \right . \implies \left \{ \begin{array}{lcl} c_1 x & = & c_2y = A\\ x^2 + y^2 & = & 2A \end{array} \right . $$ So $\nabla f \cdot \nabla g = 4x(x-c_1) + 4y(y-c_2) = 4(x^2+y^2-c_1x-c_2y) = 4(2A - A - A) = 0$