The differential equation of all circles whic | vedclass

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Question

(A) The circle passes through the origin $(0,0)$ and its centre lies on the $Y$-axis. Let the centre be $(0, k)$. Since it passes through the origin,t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The circle passes through the origin $(0,0)$ and its centre lies on the $Y$-axis. Let the centre be $(0, k)$. Since it passes through the origin,the radius is $k$.The equation of the circle is $(x-0)^2 + (y-k)^2 = k^2$.Expanding this,we get $x^2 + y^2 - 2yk + k^2 = k^2$,which simplifies to $x^2 + y^2 = 2yk$.From this,we find $k = \frac{x^2+y^2}{2y}$.Differentiating $x^2 + y^2 = 2yk$ with respect to $x$:$2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx}$.Substituting $k = \frac{x^2+y^2}{2y}$ into the equation:$2x + 2y \frac{dy}{dx} = 2 \left( \frac{x^2+y^2}{2y} \right) \frac{dy}{dx}$.$2x + 2y \frac{dy}{dx} = \left( \frac{x^2+y^2}{y} \right) \frac{dy}{dx}$.Multiplying by $y$: $2xy + 2y^2 \frac{dy}{dx} = (x^2+y^2) \frac{dy}{dx}$.Rearranging terms: $2xy = (x^2+y^2-2y^2) \frac{dy}{dx}$.$2xy = (x^2-y^2) \frac{dy}{dx}$.Thus,$(x^2-y^2) \frac{dy}{dx} - 2xy = 0$.

The differential equation of all circles which pass through the origin and whose centres lie on the $Y$-axis is

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