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General solution of following differential $\in a ^n$ differential equations of a family of circles passing through the origin and having centre on the x axis

$(a)\;x^2-y^2+2xy \frac{dy}{dx}=0 \\ (b)\;x^2-y^2+2xy \frac{dx}{dy}=0 \\ (c)\;x^2+y^2+2xy \frac{dy}{dx}=0 \\ (d)\;x^2+y^2-2xy \frac{dy}{dx} =0 $
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Let r be radius
centre (r,0)
$(x-r)^2+y^2 =r^2$
$x^2+y^2-2rx +r^2=r^2$
$2x +2y \large\frac{dy}{dx}$$-2r=0$
$x+ y \large\frac{dy}{dx}$$=r$
$x^2+y^2-2x \bigg(x +y \large\frac{dy}{dx}\bigg)$$=0$
$x^2+y^2-2x^2-2xy \large\frac{dy}{dx}$$=0$
$y^2-x^2=2xy \large\frac{dy}{dz}$
Hence d is the correct answer.
answered Feb 3, 2014 by meena.p

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