A Problem That Involves Differential Equations, Implicit Differentiation, and Tangent Lines of Circles Here is the Statement of the Problem:
Consider the family $\mathbb F$ of circles given by
$$ \mathbb F:x^2+(y-c)^2=c^2, c \in \mathbb R. $$
(a) Write down an ODE $y'=F(x,y)$ which defines the direction field of the trajectories of $\mathbb F$. Draw a sketch.
(b) Write down an ODE which defines a direction field perpendicular to the one you found in part (a). That is, find a direction field whose slope at $(x,y)$ in the phase plane is orthogonal to the slope given by $F(x,y)$. Draw a sketch. Hint: Use the fact that if $y_1$ and $y_2$ are orthogonal curves, then at the point of intersection:
$$ \frac{dy_1}{dx}\frac{dy_2}{dx}=-1.$$
(c) Find the curve through $(1,1)$ which meets every circle in the family $\mathbb F$ at an angle of $90^\circ$. Draw a sketch. Hint: Recall that the angle of intersection between two curves is defined as the angle between their tangent lines at the point of intersection.
Where I Am:
I think I've figured out everything except for part (c). I used implicit differentiation to figure out part (a), giving me:
$$ y'_1 = \frac{-x}{(y-c)}. $$
Then, naturally, the ODE for part (b) is simply:
$$ y'_2 = \frac{(y-c)}{x}.$$
Now, for part (c), perhaps I'm just not sure what's being asked. In order to find the desired curve, I should certainly consider the circle through $(1,1)$ within the family, which is a circle of radius $1$ centered at $(0,1)$. So, the line passing through that point that's tangent to that particular circle is simply $x=1$; but that line does not appear to "meet every circle in the family at an angle of $90^\circ$." Am I missing something here? 

For part (a) you have $2x+2(y-c)y'=0,$ so 
$$y'=-x/(y-c) \tag{1}$$
 as you say. However this is not the differential equation for the whole family since it still mentions the specific constant $c.$ 
From the initial relation $x^2+(y-c)^2=c^2$ you can solve for $c$ after multiplying it through to $x^2+y^2-2cy=0$ since the $c^2$ terms cancel. That gives $c=(x^2+y^2)/(2y)$ which may then be substituted for the copy of $c$ on the right side of $(1),$ and the result can be simplified if you wish. This way we get a differential equation not mentioning the constant $c$ which will hold for any of the circles.
[note it's not that your formula is wrong, it's just that (in most diff eq books I know of) the differential equation of a family of curves having a parameter usually means one somehow solves it and then expresses the parameter in terms of the original variables and plugs that in, to arrive at an equation not mentioning the parameter.]
Added: On simplifying things the above answer gets to $y'=(2xy)/(x^2-y^2).$ Note that where the denominator is zero here is on the lines $y=\pm x$ and the family of circles are all those centered on the $y$ axis and passing through the origin, and their tangents are indeed vertical on the lines $y = \pm x.$
If one does the same thing with the circles $(x-c)^2+y^2=c^2,$ which are the circles centered on the $x$ axis passing through the origin, one gets in this case that $y'=-(x^2-y^2)/(2xy),$ which is the negative reciprocal of the other circle family, so that each circle of the second family is orthogonal to any of the first family where these meet. [I worked this all out5 once, and there were some involved steps getting to the orthogonal family of the other collection of circles...]
Finally for part (c) it looks like one wants the circle of the second family (center on $x$ axis, passing through the origin) which passes through $(1,1).$ This circle has center $(1,0)$ and radius $1,$ equation $(x-1)^2+y^2=1^2.$
