The differential equation of the system of circles touching the y-axis at the
origin is
(a) $x^2+y^2-2xy\dfrac{dy}{dx}=0$
(b) $x^2+y^2+2xy\dfrac{dy}{dx}=0$
(c) $x^2-y^2-2xy\dfrac{dy}{dx}=0$
(d) $x^2-y^2+2xy\dfrac{dy}{dx}=0$
The equation of the system of circles touching the y-axis at the origin is
$(x\pm r)^2+y^2=r^2$
$x^2+r^2\pm2xr+y^2=r^2$ .. (1)
$x^2\pm2xr+y^2=0$
Differentiating w.r.t x, $2x\pm2r+2y\dfrac{dy}{dx}=0$
$\dfrac{dy}{dx}=\dfrac{\mp r-x}{y}$
Putting this back in any of the above options, does not give (1) back.
Best Answer
Let's put it back to (d): $$x^2-y^2+2xy\dfrac{dy}{dx} = x^2-y^2+2xy\dfrac{\mp r-x}{y}=x^2-y^2-2x^2\mp 2xr=-(x^2+y^2\pm2xr)=0$$ which is exactly the one below (1).