# What would the second degree equation be?

Find a second degree differential equation describing family of circles: $$(x-a)^2 + (y-b)^2 = b^2.$$ I differentiated once and obtained: $$x'(t) (x(t) - a) + y'(t) (y(t) - b) = 0,$$ then differentiated for the second time and got: $$(x'(t))^2 + x(t)x''(t) - ax''(t) + (y'(t))^2 + y(t)y'(t) - by''(t) =0.$$ And...what? It doesn't make my work any easier, it doesn't simplify my first equation really... How to solve it? I would really appreciate any help.

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Are you supposed to find a differential equation of the form $F(x'(t),y'(t),t)=0$? Or of the form $F(y'(x),x)=0$? –  Avi Steiner Mar 3 '13 at 19:01
You want to get rid of the constants $a$ and $b$ in your resulting second order ODE by combining the derivatives. –  vonbrand Mar 3 '13 at 19:01

Why don't you treat $y$ as a function of $x$ and differentiate three times with respect to $x$? Differentiating once gives $(x - a) + y'(y - b) = 0$. Differentiating again gives $$1 + y''(y - b) + (y')^2 = 0 \tag{1} \\$$ $$y - b = -\frac{1 + (y')^2}{y''}. \tag{2}$$ Differentiating $(1)$ gives the final equation we need $$y'''(y - b) + y''y' + 2y'y'' = 0$$ $$y - b = -\frac{3y'y''}{y'''}. \tag{3}$$ We can equate $(2)$ and $(3)$ to get $$\frac{1 + (y')^2}{y''} = \frac{3y'y''}{y'''} \\ y'''(1 + (y')^2) = 3y'(y'')^2 \\ y'''(1 + (y')^2) - 3y'(y'')^2 = 0$$ whichever form you prefer.
@Did Of course not. But if you want the differential equation for the family of all circles (which I did) then there are three constant: $a, b, r$, where $r$ is the radius (I am assuming that the equation written in the question has a typo for the radius squared). Thus, you can't have a second order equation. If the question did not have a typo, then it represents a smaller family of circles and to find a differential equation which are only satisfied by those circles would require a different approach. –  Pratyush Sarkar Mar 6 '13 at 1:00
$$x''(t)=-y'(t)\qquad y''(t)=x'(t)$$ $$\text{or, equivalently,}$$ $$(x''(t)+y'(t))^2+(y''(t)-x'(t))^2=0$$