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Question

(C) The equation of a circle with center $(h, 0)$ on the $x$-axis and radius $h$ (since it touches the $y$-axis at the origin) is $(x - h)^2 + y^2 = h

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of a circle with center $(h, 0)$ on the $x$-axis and radius $h$ (since it touches the $y$-axis at the origin) is $(x - h)^2 + y^2 = h^2$.Expanding this,we get $x^2 - 2xh + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 = 2xh$.To eliminate the arbitrary constant $h$,we differentiate with respect to $x$:$2x + 2y \frac{dy}{dx} = 2h$.Substituting $h = x + y \frac{dy}{dx}$ into the original equation $x^2 + y^2 = 2x(h)$:$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$.$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$.$y^2 - x^2 = 2xy \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation of all circles having center on the $x$-axis and touching the $y$-axis at the origin is:

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