Tenenbaum and Pollard, Ordinary Differential Equations, problem 1.4.29, what am I missing? Tenenbaum and Pollard's "Ordinary Differential Equations," chapter 1, section 4, problem 29 asks for a differential equation whose solution is "a family of straight lines that are tangent to the circle $x^2 + y^2 = c^2$, where $c$ is a constant."
Since the solutions will be lines, I start with the formula $y = m x + b$, and since the line is determined by a single parameter (the point on the circle to which the line is tangent) I expect the differential equation to be of order one.  Differentiating, I get $y' = m$, so $y = y' x + b$.
So now, I need an equation for $b$.  The solution given in the text is $y = x y' \pm c \sqrt{(y')^2 + 1}$, implying $b = \pm c \sqrt{(y')^2 + 1}$, but try as I might I have been unable to derive this formula for $b$.  I'm sure I'm missing something simple.

I'll assume the point $P=(x,y)$ lies on the circle $x^2+y^2=c^2$ in the first quadrant. The slope of the tangent at $P$ is $y'$ as you say. You need to express the $y$ intercept. 
Extend the tangent line until it meets the $x$ axis $A$ and the $y$ axis at $B$, and call the origin $O$. Then the two triangles $APO$ and $OPB$ are similar. From this you can get the y intercept, which is the point $(0,OB)$ by use of
$OB=OP*(AB/OA)=OP*sqrt([OA^2+OB^2]/OA^2)=OP*sqrt(1+[OB/OA]^2)$. And $y'=-OB/OA$, being the slope of the line joining $A$ to $B$ lying respectively on the $x$ and $y$ axes. Finally the $OP$ here is the constant $c$, the circle's radius.
