The differential equation of the family of ci | vedclass

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Question

(A) The general equation of a circle passing through $(0, 2)$ and $(0, -2)$ is given by the family of circles $x^{2} + y^{2} + 2gx + 2fy + c = 0$.Sinc

Vedclass · Accepted Answer

$2 x y \frac{d y}{d x} + (x^{2} - y^{2} + 4) = 0$

Vedclass · Answer

(A) The general equation of a circle passing through $(0, 2)$ and $(0, -2)$ is given by the family of circles $x^{2} + y^{2} + 2gx + 2fy + c = 0$.Since the points $(0, 2)$ and $(0, -2)$ lie on the circle,we have $4 + 4f + c = 0$ and $4 - 4f + c = 0$,which implies $f = 0$ and $c = -4$.Thus,the equation of the family of circles is $x^{2} + y^{2} + 2gx - 4 = 0$,which can be written as $x^{2} + y^{2} - 4 + 2gx = 0$.Dividing by $x$ (assuming $x 
eq 0$),we get $\frac{x^{2} + y^{2} - 4}{x} + 2g = 0$.Differentiating with respect to $x$,we get $\frac{d}{dx} \left( \frac{x^{2} + y^{2} - 4}{x} ight) = 0$.Using the quotient rule,$\frac{x(2x + 2y \frac{dy}{dx}) - (x^{2} + y^{2} - 4)(1)}{x^{2}} = 0$.This simplifies to $2x^{2} + 2xy \frac{dy}{dx} - x^{2} - y^{2} + 4 = 0$.Rearranging the terms,we get $2xy \frac{dy}{dx} + x^{2} - y^{2} + 4 = 0$.

The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is

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