Step 1:

The given equation is $y^2-2ay+x^2=a^2$--------(1)

Differentiating with respect to $x$ we get,

$2yy'-2ay'+2x=0$

$\Rightarrow yy'-ay'+x=0$

$\therefore \large\frac{yy'+x}{y'}$$=a$

Step 2:

Now substitute for $a$ in equation(1)

$y^2-2\big(\large\frac{yy'+x}{y'}\big)$$+x^2=\big(\large\frac{yy'+x}{y'}\big)^2$

$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)^2$$+2\big(\large\frac{yy'+x}{y'}\big)$

$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)\bigg[\large\frac{yy'+x}{y'}$$+2\bigg]$

$\Rightarrow x^2+y^2=\big(\large\frac{yy'+x}{y'}\big)\bigg[\large\frac{yy'+x+2y'}{y'}\bigg]$

$(y')^2[x^2+y^2]-(yy'+x)(yy'+x+2y')=0$

$\therefore x(y')^2-2y(y')^2-2xy'-x^2=0$

This is the required equation.