An elementary (and not particularly smart) approach could be:
Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$
Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.
You know that $y'(0)=f(0,y_0)$.
Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$
Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.
Keep deriving
$$y^{(3)}(x)=\\=
\dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x))
$$
Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.
The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.
Assume the equation of the line is $ax+by=c$ and the point is $(x_0,y_0).$ A vector which gives the direction of the line is $(b,-a).$ Since the line and the circle are tangent at $(x_0,y_0)$ the center is on the line which is perpendicular to the given one. The vector which gives the direction of the perpendicular line is $(a,b).$ Thus, the equation of all possible circles is
$$(x-(x_0\pm ra))^2+(y-(y_0\pm rb))^2=r^2(a^2+b^2)$$
where $r\in (0,\infty).$
We can assume without lost of generality that $a^2+b^2=1.$ In such a case we have that the equations of all possible circles are
$$(x-(x_0\pm ra))^2+(y-(y_0\pm rb))^2=r^2$$
where $r\in (0,\infty)$ is the radius.
Best Answer
then circles are given by $$(x-a)^2 + (y-a)^2 = 1, a \text{ arbitrary}. \tag 1 $$ differentiating $(1)$ with respect to $x,$ we have $$(x-a) +(y-a)y' = 0\tag 2 $$ solving $(2)$ for $a,$ we get $$a = \frac{x+yy'}{1+y'}, \quad x - a=\frac{(x-y)y'}{1+y'}, \quad y - a =\frac{y-x}{1+y'}\tag 3 $$
subbing $(3)$ in $(1)$ gives you $$(x-y)^2\left(y'^2+1\right) = (1+y')^2 $$