The differential equation of the family of circles passing through the origin and having centre at the line $$y = x$$ is :

Family of circles passing through the origin with centre on $$y = x$$.

General equation of such a circle.

Centre at $$(a, a)$$ (since it's on $$y = x$$). The circle passes through origin, so radius $$= \sqrt{a^2 + a^2} = a\sqrt{2}$$.

Equation: $$(x - a)^2 + (y - a)^2 = 2a^2$$

$$ x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = 2a^2 $$ $$ x^2 + y^2 - 2a(x + y) = 0 $$

So: $$a = \frac{x^2 + y^2}{2(x + y)}$$ ... (parameter to eliminate)

Differentiate.

$$ 2x + 2yy' - 2a(1 + y') = 0 $$ $$ a = \frac{x + yy'}{1 + y'} $$

Equate both expressions for $$a$$.

$$ \frac{x^2 + y^2}{2(x + y)} = \frac{x + yy'}{1 + y'} $$ $$ (x^2 + y^2)(1 + y') = 2(x + y)(x + yy') $$

Expand LHS: $$(x^2 + y^2) + (x^2 + y^2)y'$$

Expand RHS: $$2(x^2 + xyy' + xy + y^2y') = 2x^2 + 2xyy' + 2xy + 2y^2y'$$

Collecting terms:

$$ (x^2 + y^2) + (x^2 + y^2)y' = 2x^2 + 2xy + (2xy + 2y^2)y' $$

Terms without $$y'$$: $$x^2 + y^2 - 2x^2 - 2xy = y^2 - x^2 - 2xy$$

Terms with $$y'$$: $$(x^2 + y^2 - 2xy - 2y^2)y' = (x^2 - y^2 - 2xy)y'$$

$$ (y^2 - x^2 - 2xy) + (x^2 - y^2 - 2xy)y' = 0 $$ $$ (x^2 - y^2 + 2xy) = (x^2 - y^2 - 2xy)y' $$

Writing as: $$(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy$$

The correct answer is Option (1): $$(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy$$.