Find the diff equation of family of circles with center on the line y= -x and passing through the origin.
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$(x - h)^2 + (y - k)^2 = r^2$
$(x - h)^2 + (y + h)^2 = 2h^2$
$2(x - h) + 2(y + h)y' = 0$
$(x - h) + (y + h)y' = 0$
$-h(1 - y') = -(x + yy')$
$h = \dfrac{x + yy'}{1 - y'}$
$\left( x - \dfrac{x + yy'}{1 - y'} \right)^2 + \left( y + \dfrac{x + yy'}{1 - y'} \right)^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$
$\left[ \dfrac{(x - xy') - (x + yy')}{1 - y'} \right]^2 + \left[ \dfrac{(y - yy') + (x + yy')}{1 - y'} \right]^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$
$\left[ (x - xy') - (x + yy') \right]^2 + \left[ (y - yy') + (x + yy') \right]^2 = 2\left( x + yy' \right)^2$
$(x + y)^2 (y')^2 + (x + y)^2 = 2(x + yy')^2$
$$\begin{eqnarray} (x-h)^2 + (y+h)^2 &=& 2h^2\\ x^2 + y^2 - 2hx - 2hy &=& 0\\ \dfrac{x^2 + y^2}{x+y} &=& 2h\\ \dfrac{(2x + 2yy')(x+y) - (x^2+y^2)(1+y')}{(x+y)^2} &=& 0\\ (2x+2yy')(x+y) - (x^2+y^2)(1+y') &=& 0\\ (2xy+2y^2 -x^2 - y^2)y' + (2x^2 +2xy - x^2 - y^2) &=& 0\\ (-x^2 + 2xy + y^2)y' + (x^2 + 2xy - y^2) &=& 0\\ (x^2 - 2xy - y^2)dy - (x^2 + 2xy - y^2)dx &=& 0 \leftarrow Answer \end{eqnarray}$$
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$(x - h)^2 + (y - k)^2 = r^2$
$(x - h)^2 + (y + h)^2 = 2h^2$
$2(x - h) + 2(y + h)y' = 0$
$(x - h) + (y + h)y' = 0$
$-h(1 - y') = -(x + yy')$
$h = \dfrac{x + yy'}{1 - y'}$
$\left( x - \dfrac{x + yy'}{1 - y'} \right)^2 + \left( y + \dfrac{x + yy'}{1 - y'} \right)^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$
$\left[ \dfrac{(x - xy') - (x + yy')}{1 - y'} \right]^2 + \left[ \dfrac{(y - yy') + (x + yy')}{1 - y'} \right]^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$
$\left[ (x - xy') - (x + yy') \right]^2 + \left[ (y - yy') + (x + yy') \right]^2 = 2\left( x + yy' \right)^2$
$(x + y)^2 (y')^2 + (x + y)^2 = 2(x + yy')^2$
$$\begin{eqnarray}
(x-h)^2 + (y+h)^2 &=& 2h^2\\
x^2 + y^2 - 2hx - 2hy &=& 0\\
\dfrac{x^2 + y^2}{x+y} &=& 2h\\
\dfrac{(2x + 2yy')(x+y) - (x^2+y^2)(1+y')}{(x+y)^2} &=& 0\\
(2x+2yy')(x+y) - (x^2+y^2)(1+y') &=& 0\\
(2xy+2y^2 -x^2 - y^2)y' + (2x^2 +2xy - x^2 - y^2) &=& 0\\
(-x^2 + 2xy + y^2)y' + (x^2 + 2xy - y^2) &=& 0\\
(x^2 - 2xy - y^2)dy - (x^2 + 2xy - y^2)dx &=& 0 \leftarrow Answer
\end{eqnarray}$$
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