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Find a first-order differential equation whose integral curves are all circles through (1,1) and (-1,-1)

by RoRi

Find a first-order differential equation whose integral curves consist of all circles through the points (1,1) and (-1,-1).

Circles going through both the points (1,1) and (-1,-1) must have their center on the line y = -x, say at (C,-C). The radius is given by

    \[ r^2 = (C-1)^2 +(-C-1)^2 = 2(C^2+1). \]

Therefore we have the equation

    \[ (x-C)^2 + (y+C)^2 = 2(C^2+1). \]

Differentiating both sides with respect to x,

    \[ (x-C)^2 + (y+C)^2 = 2(C^2 + 1) \quad \implies \quad 2(x-C) + 2(y+C)y' = 0. \]

From the original equation we can solve for the constant,

    \[ (x-C)^2 + (y+C)^2 = 2(C^2 + 1) \quad \implies \quad C = \frac{2-x^2-y^2}{2y-2x}. \]

Therefore,

    \begin{align*} && 2(x-C) + 2(y+C)y' &= 0 \\ \implies && x - \frac{2-x^2-y^2}{2y-2x} + y'y + \left( \frac{2-x^2-y^2}{2y-2x} \right) y' &= 0 \\ \implies && 2xy - 2x^2 - 2 + x^2 + y^2 + 2y^2 y' - 2xyy' + 2y' - x^2 y' -y^2y' &= 0\\ \implies && y' (x^2 - y^2 + 2xy-2) - y^2 - 2xy + x^2 + 2 &= 0. \end{align*}

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