Equation of circle touching a parabola Suppose we have a parabola $y^2=4x$  . Now, how to write equation of circle touching parabola at $(4,4)$ and passing thru focus?
I know that for this parabola focus will lie at $(1,0)$ so we may assume general equation of circle and satisfy the points in it . hence
$$x^2 +y^2 + 2gx + 2fy +c =0$$ should be equation of general circle but on satisfying given points we get 2 equations and 3 variables . I think I am missing something, what to do?

A non standard solution
Take a second point on the parabola very close to $(4,4)$: $(4+\epsilon,4+2\epsilon)$ agrees to the first order (plugging in the equation of the parabola, $16+8\epsilon+\epsilon^2\approx16+8\epsilon$).
Then solve the circle by the three points
$$\begin{align}
(x-4)^2&+(y-4)^2&=r^2,\\
(x-4-2\epsilon)^2&+(y-4-\epsilon)^2&=r^2,\\
(x-1)^2&+y^2&=r^2
\end{align}$$
Subtracting $(1)-(2)$* and $(1)-(3)$* and ignoring $\epsilon^2$:
$$2\epsilon(2x-8)+\epsilon(2y-8)=0,\\3(2x-5)+4(2y-4)=0.$$
Hence
$$x=\frac{13}2,y=-1.$$

$$\text{* }(x-x_0)^2+(y-y_0)^2-(x-x_1)^2-(y-y_1)^2\\
=(x_1-x_0)(2x-x_0-x_1)+(y_1-y_0)(2y-y_0-y_1)$$

