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?
the tangent to the parabola at $(4, 4)$ has slope $1/2$ so the radius has slope $-2.$ let the center of the circle touching the parabola $y^2 = 4x$ at $(4,4)$ be 
$x = 4 + t, y = 4 - 2t$. now equating the radius $$5t^2 = (3+t)^2 + (4-2t)^2$$
you can find $t$ which will give you the center and the radius of the circle.
If what you mean by "touching" is "intersect", then this is the solution. Your circle is defined by its center $(x_c,y_c)$ and its radius $R$:
$$(x-x_c)^2+(y-y_c)^2=R^2$$
Now, you want all three points $(1,0)$, $(4,4)$ and $(4,-4)$ to be on the circle, so you hav to solve the system:
\begin{cases}(1-x_c)^2+(0-y_c)^2=R^2\\(4-x_c)^2+(4-y_c)^2=R^2\\(4-x_c)^2+(-4-y_c)^2=R^2\end{cases}
which gives
$$x_c=\dfrac{31}{6},\ \ y_c=0,\ \ R=\dfrac{25}{6}$$
or
$$\left(x-\dfrac{31}{6}\right)^2+y_c^2=\dfrac{625}{36}$$