The differential equation of all circles whic | vedclass

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Question

(A) The equation of a circle passing through the origin with its center on the $y$-axis is given by ${x^2} + {(y - a)^2} = {a^2}$,where $(0, a)$ is th

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a circle passing through the origin with its center on the $y$-axis is given by ${x^2} + {(y - a)^2} = {a^2}$,where $(0, a)$ is the center.This simplifies to ${x^2} + {y^2} - 2ay = 0$.Differentiating both sides with respect to $x$,we get $2x + 2y\frac{{dy}}{{dx}} - 2a\frac{{dy}}{{dx}} = 0$.From this,we find $a\frac{{dy}}{{dx}} = x + y\frac{{dy}}{{dx}}$,so $a = \frac{{x + y\frac{{dy}}{{dx}}}}{\frac{{dy}}{{dx}}} = x\frac{{dx}}{{dy}} + y$.Substituting the value of $2a$ into the original equation ${x^2} + {y^2} - 2ay = 0$:${x^2} + {y^2} - 2y(x\frac{{dx}}{{dy}} + y) = 0$.${x^2} + {y^2} - 2xy\frac{{dx}}{{dy}} - 2{y^2} = 0$.${x^2} - {y^2} - 2xy\frac{{dx}}{{dy}} = 0$.Multiplying by $\frac{{dy}}{{dx}}$,we get $({x^2} - {y^2})\frac{{dy}}{{dx}} - 2xy = 0$.

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

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