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

We need the differential equation of the family of circles passing through $$(0, 2)$$ and $$(0, -2)$$.

The general circle equation is $$x^2 + y^2 + 2gx + 2fy + c = 0$$. Substituting $$(0, 2)$$: $$4 + 4f + c = 0$$. Substituting $$(0, -2)$$: $$4 - 4f + c = 0$$.

Adding these two equations gives $$8 + 2c = 0$$, so $$c = -4$$. Subtracting gives $$8f = 0$$, so $$f = 0$$.

The family of circles is therefore $$x^2 + y^2 + 2gx - 4 = 0$$, with $$g$$ as the only parameter to be eliminated.

Differentiating with respect to $$x$$: $$2x + 2y\frac{dy}{dx} + 2g = 0$$, which gives $$g = -x - y\frac{dy}{dx}$$.

Substituting this back into the circle equation:

$$x^2 + y^2 + 2\left(-x - y\frac{dy}{dx}\right)x - 4 = 0$$

$$x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} - 4 = 0$$

Simplifying: $$-x^2 + y^2 - 2xy\frac{dy}{dx} - 4 = 0$$. Multiplying through by $$-1$$:

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

Hence, the correct answer is Option A.