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(A) The equation of a family of circles touching the $Y$-axis at the origin $(0,0)$ is given by $(x-a)^2 + y^2 = a^2$,where $a$ is the radius of the c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a family of circles touching the $Y$-axis at the origin $(0,0)$ is given by $(x-a)^2 + y^2 = a^2$,where $a$ is the radius of the circle and $(a,0)$ is the center.Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$ $(i)$.Differentiating both sides with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2a$.Thus,$a = x + y \frac{dy}{dx}$ $(ii)$.Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$$2xy \frac{dy}{dx} = y^2 - x^2$$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation corresponding to the family of circles in the plane touching the $Y$-axis at the origin is:

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