The question is: Consider the family $F$ of circles in the $xy$ plane, $(x-c)^2+y^2=c^2$ tangent to the $y$ axis at the origin. Find a differential equation that is satisfied by the family of curves orthogonal to $F$.
My thinking: Since the implicit equation represents the level sets of the function $$ f(x,y)=c^2=(x-c)^2+y^2 $$ The gradient of the function $f$ will be perpendicular to its level sets, and therefore orthogonal to the family of curves $F$. This yields $$ \nabla f(x,y)=(0,0)=(2x-2c,2y)\Rightarrow \left(x-\frac{x^2+y^2}{2x},y\right)=(0,0)\\ \Rightarrow \left(\frac{x^2-y^2}{2x},y\right)=(0,0) $$ So we have in differential form $$ \frac{x^2-y^2}{2x}dx+ydy=0\Rightarrow \frac{y^2-x^2}{2xy}=\frac{dy}{dx} $$ But the answer is the negative reciprocal, or perpendicular vector to this one. Why? I assume my reasoning was flawed in the first step, when i took the gradient of $f$ to be perpindicular to the family $F$, but I don't see why.