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(B) Since the circles touch the $y$-axis at the origin,their centers must lie on the $x$-axis. Let the center be $(h, 0)$ and the radius be $h$. The e

Vedclass · Accepted Answer

$x^2-y^2+2xy\frac{dy}{dx}=0$

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(B) Since the circles touch the $y$-axis at the origin,their centers must lie on the $x$-axis. Let the center be $(h, 0)$ and the radius be $h$. The equation of the circle is $(x-h)^2 + y^2 = h^2$,which simplifies to $x^2 - 2hx + h^2 + y^2 = h^2$,or $x^2 + y^2 - 2hx = 0$ $(1)$. Differentiating both sides with respect to $x$,we get $2x + 2y\frac{dy}{dx} - 2h = 0$,which simplifies to $h = x + y\frac{dy}{dx}$. Substituting this value of $h$ into equation $(1)$,we get $x^2 + y^2 - 2(x + y\frac{dy}{dx})x = 0$. Expanding this,we get $x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} = 0$. Rearranging the terms,we obtain $y^2 - x^2 - 2xy\frac{dy}{dx} = 0$,which is equivalent to $x^2 - y^2 + 2xy\frac{dy}{dx} = 0$.

The differential equation of the family of circles touching $y$-axis at the origin is

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