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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.

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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.

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Thank you so much. I see it now. Did you mean to note that ABO is similar to OPB instead of that APO is similar to OPB? All three triangles have the same angles, but the length AB does not appear in either APO or OPB. –  Aaron Golden Oct 14 '12 at 8:20
    
Yes I wrote hastily. The main thing is the first statement gets OB in terms of other lengths. To say OB=OP*(AB/OA) comes from a few combinations, one being OB/OP=AB/OA. This is a comparison between the "upper" small triangle and the "big triangle" OAB. By the time I wrote it I had too many diagrams on pieces of paper and I wasn't careful enough. –  coffeemath Oct 14 '12 at 8:29
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