You may know that, among the different ways to write the equation of a straight line, one of them is :
$$\tag{1}\dfrac{x}{x_0}+\dfrac{y}{y_0}=1$$
where $x_0$ is the abscissa of the intercept with the $x$ axis, and $y_0$ is the ordinate of the intercept with the $y$ axis.
(check for example that if $y=y_0$ then $x=0$ ; the same thing when $x=x_0$).
Then, as $x_0+y_0=k$, one may write (1) under the form:
$$\tag{2}\dfrac{x}{k-y_0}+\dfrac{y}{y_0}=1 \ \iff \ y = y_0(1 - \dfrac{x}{k-y_0}) \ \iff \ $$
$$\tag{3}y = \left(\dfrac{y_0}{y_0-k}\right)x+y_0$$
which depends on the single parameter $x_0$.
Now, if, in the left hand side of differential equation
$$\tag{4} \ (xy'-y)(y'-1)+ky'=0$$
we replace $y$ by (3) and $y'$ by $\left(\dfrac{y_0}{y_0-k}\right)$, it is easy to check that we obtain a right hand side in (4) which is $0$ (variable $y_0$ vanishes in the computation).
In a converse way, one could ask whether all solutions of the given differential equation are straight lines with equation (3).
@amd's answer gave me the hint.
First, if we take $\lambda$ to be $-1$ then we get a non homogeneous equation of first degree($x+y=4$) which would surely represent a straight line passing through the points of intersection of the two curves. Now if we somehow homogenize the equation of the second circle withe the help of the equation of the line, i.e $x+y=4$, to a second degree equation, then this equation would surely represent a pair of straight lines through the origin, passing through the points of intersection of the two curves.
I did this by writing $1=\frac{x+y}{4}$ and so multiplying this thing in the RHS with all the terms of first degree in the equation of the second curve and the square of this to the term of zeroth degree, i.e the constant $+14$.
This gives the equation of the pair of straight lines:
$$x^2+y^2-6x\left(\frac{x+y}{4}\right)-6y\left(\frac{x+y}{4}\right)+14\left(\frac{x+y}{4}\right)^2=0$$
Best Answer
Here is an attempt to find a differential equation whose solutions are precisely the lines at distance $p$ from the origin.
We follow the question as far as $$y'=-\cot A$$ and then we try to eliminate $A$ in favor of $x,y,p$. We go back to $$x\cos A+y\sin A=p$$ Dividing through by $\sqrt{x^2+y^2}$, letting $\theta=\arctan(y/x)$, and using $$\cos(r-s)=\cos r\cos s+\sin r\sin s$$ we get $$\cos(A-\theta)={p\over\sqrt{x^2+y^2}}$$ Solving for $A$ we get $$A=\theta+\arccos{p\over\sqrt{x^2+y^2}}$$ so we have the differential equation $$y'=-\cot\left(\arctan(y/x)+{p\over\sqrt{x^2+y^2}}\right)$$ I managed to "simplify" this to $$y'={y\tan{p\over\sqrt{x^2+y^2}}-x\over y+x\tan{p\over\sqrt{x^2+y^2}}}$$
There must be a better way.
EDIT: Maybe it looks a little better as $$xy'+y(1+(y')^2)=p\sqrt{1+(y')^2}$$ or maybe not.