The differential equation of the family of circles passing through the origin and having centre at the line $y=x$ is:
$(x^2-y^2+2xy)\,dx=(x^2-y^2-2xy)\,dy$
$(x^2+y^2+2xy)\,dx=(x^2+y^2-2xy)\,dy$
$(x^2+y^2-2xy)\,dx=(x^2+y^2+2xy)\,dy$
$(x^2-y^2+2xy)\,dx=(x^2-y^2+2xy)\,dy$
Step-by-Step Solution
Key Concept: Circle: $x^2+y^2+gx+gy=0$ (centre $(-g/2,-g/2)$ on $y=x$, passes through origin). Differentiate, eliminate $g$.
$(x^2-y^2+2xy)dx=(x^2-y^2-2xy)dy$.
Correct Answer: 1