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

Let the centre of the circle be \((a,a)\), since the centre lies on the line \(y=x\).

Since the circle passes through the origin \((0,0)\),

its radius is the distance between \((a,a)\) and \((0,0)\).

So, \(r^2=a^2+a^2=2a^2\).

Therefore, the equation of the circle is \((x-a)^2+(y-a)^2=2a^2\).

\(x^2+y^2-2a(x+y)=0\).

\(2x+2y\frac{dy}{dx}-2a\left(1+\frac{dy}{dx}\right)=0\)

Let \(\frac{dy}{dx}=y'\).

So, \(2x+2yy'-2a(1+y')=0\).

Hence, \(a=\frac{x+yy'}{1+y'}\).

From the circle equation,

\(x^2+y^2=2a(x+y)\).

Substitute \(a=\frac{x+yy'}{1+y'}\).

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

\((x^2+y^2)(1+y')=2(x+y)(x+yy')\).

\(x^2+y^2+x^2y'+y^2y'=2x^2+2xy+2xyy'+2y^2y'\)

So, \((x^2-2xy-y^2)y'+y^2-x^2-2xy=0\).

Therefore, \((x^2-2xy-y^2)\frac{dy}{dx}=x^2+2xy-y^2\).