The differential equation of all circles whic | vedclass

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Question

(A) Let the centre of the circle be $(0, k)$. Since the circle passes through the origin $(0, 0)$,the radius of the circle is $r = \sqrt{(0-0)^2 + (k-

Vedclass · Accepted Answer

$\left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0$

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(A) Let the centre of the circle be $(0, k)$. Since the circle passes through the origin $(0, 0)$,the radius of the circle is $r = \sqrt{(0-0)^2 + (k-0)^2} = |k|$.The equation of the circle is $(x-0)^2 + (y-k)^2 = k^2$.Expanding this,we get $x^2 + y^2 - 2ky + k^2 = k^2$,which simplifies to $x^2 + y^2 = 2ky$.To eliminate the arbitrary constant $k$,we differentiate with respect to $x$:$2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx}$.From the circle equation,$k = \frac{x^2+y^2}{2y}$.Substituting this value of $k$ into the differentiated equation:$2x + 2y \frac{dy}{dx} = 2 \left( \frac{x^2+y^2}{2y} \right) \frac{dy}{dx}$.$2x + 2y \frac{dy}{dx} = \frac{x^2+y^2}{y} \frac{dy}{dx}$.Multiplying by $y$:$2xy + 2y^2 \frac{dy}{dx} = (x^2+y^2) \frac{dy}{dx}$.Rearranging the terms:$(x^2+y^2) \frac{dy}{dx} - 2y^2 \frac{dy}{dx} = 2xy$.$(x^2-y^2) \frac{dy}{dx} = 2xy$.Thus,$(x^2-y^2) \frac{dy}{dx} - 2xy = 0$.

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

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