The differential equation of the family of ci | vedclass

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(B) Since the circle touches the $y$-axis at the origin,its center must lie on the $x$-axis. Let the center be $(a, 0)$ and the radius be $a$.The equa

Vedclass · Accepted Answer

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

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(B) Since the circle touches the $y$-axis at the origin,its center must lie on the $x$-axis. Let the center be $(a, 0)$ and the radius be $a$.The equation of the circle is $(x-a)^2 + (y-0)^2 = a^2$.Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$ ... $(i)$.Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx}$ ... $(ii)$.Substituting the value of $2a$ from $(ii)$ into $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$.This simplifies to $x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.Thus,the differential equation is $y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,or $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

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

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