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(D) The general equation of a circle with center $(a, 0)$ on the $X$-axis and passing through the origin $(0, 0)$ is given by $(x-a)^2 + y^2 = a^2$. E

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The general equation of a circle with center $(a, 0)$ on the $X$-axis and passing through the origin $(0, 0)$ is given by $(x-a)^2 + y^2 = a^2$. Expanding this, we get $x^2 - 2ax + a^2 + y^2 = a^2$, which simplifies to $x^2 + y^2 - 2ax = 0$. To eliminate the arbitrary constant $a$, we differentiate with respect to $x$: $2x + 2y\frac{dy}{dx} - 2a = 0$. This gives $a = x + y\frac{dy}{dx}$. Substituting this value of $a$ back into the equation $x^2 + y^2 = 2ax$, we get $x^2 + y^2 = 2x(x + y\frac{dy}{dx})$. Expanding the right side, $x^2 + y^2 = 2x^2 + 2xy\frac{dy}{dx}$. Rearranging the terms, we get $y^2 - x^2 - 2xy\frac{dy}{dx} = 0$.

The differential equation corresponding to the family of circles having centres on the $X$-axis and passing through the origin is

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