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: 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.
A: The center lies at the intersection of the perpendicular to the tangent and the bisector of the two known points.
$$2(x-4)+(y-4)=0$$
$$(4-1)(x-\frac{4+1}2)+(4-0)(y-\frac{4+0}2)=0.$$
Hence
$$x=\frac{13}2,y=-1.$$
A: 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
$$x=\frac{13}2,y=-1.$$

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

A: 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
$$\left(x-\dfrac{31}{6}\right)^2+y_c^2=\dfrac{625}{36}$$
