The orthogonal trajectories of the family of circles given by x^2+y^2…

The equation of the family of circles is

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

Differentiating w.r.t. $x$, we get

$2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0 \Rightarrow a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}$

Putting this value of $a$ in (i), we get

$x^2+y^2-2 y\left\{\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}\right\}=0 \Rightarrow x^2-y^2-2 x y \frac{d x}{d y}=0$       .....(ii)

This is the differential equation of the family of circles given by (i).

The differential equation representing the family of orthogonal trajectories of (i) is obtained by replacing $\frac{d y}{d x} b y-\frac{d x}{d y}$ in (ii). So, the differential equation of the orthogonal trajectories is 

$x^2-y^2+2 x y \frac{d y}{d x}=0$

$\Rightarrow \left(x^2-y^2\right) d x+2 x y d y=0$

$\Rightarrow 2 x y d y-y^2 d x=-x^2 d x$

$\Rightarrow \frac{x d\left(y^2\right)-y^2 d x}{x^2}=-d x$

$\Rightarrow d\left(\frac{y^2}{x}\right)=-d x$

$\Rightarrow \frac{y^2}{x}=-x+2 k \Rightarrow x^2+y^2-2 k x=0$

This is the required family of orthogonal trajectories.