Finding Equation of Circle from Tangent Line Slope? 
Two circles of radius 4 are tangent to the graph of 
  y^2 = 4x
   at the point 
  (1, 2).
   Find equations of these two circles. (Enter your answers as a comma-separated list.)

Ok, so I found the slope of y^2 = 4x @ (1,2 )by dy/dx implicitly. The answer I got was 1.

2 / y | (1 , 2) | y = x + 1

How would I proceed from here? I know how to find the tangent line from a circle and a given point, but how would I do the opposite?
 A: The distance between the center of the circle to the touching point should be equal to 4 and perpendicular to it.

The equation of the perpendicular line will be:
\begin{equation}g(x)=-1x+b\end{equation}
\begin{equation}g(1)=-1*1+b\end{equation}
\begin{equation}2=-1*1+b\end{equation}
\begin{equation}b=2+1\end{equation}
\begin{equation}b=3\end{equation} 
\begin{equation}g(x)=-x+3\end{equation}
The distance between the center of the circle and the point (1,2) is exactly 4 (the radius).
Using the formula of distance:
$$ (1-x_c)^2+(2-y_c)^2=4^2  $$
Where $ x_c $ and $ y_c$ are the coordinates of the center of the circle. Since the slope of the perpendicular line is -1 it means that the ratio between $\bigtriangleup x$ and $\bigtriangleup y$ is 1. Thus $1-x_c=2-y_c$. Let $a$ be $1-x_c and 2-y_c$, then
$$ a^2+a^2=4^2  $$
$$ 2a^2=4^2 $$
$$ a^2=8 $$
$$ a=\pm2\surd2$$
Then $1-x_c=\pm2\surd2$ and $2-y_c=\pm2\surd2$.
$$ 1-x_c=\pm2\surd2 $$ 
$$ x_c=1\pm2\surd2 $$ 
Same with $2-y_c=\pm2\surd2$. 
$$ y_c=2\pm2\surd2 $$
The final equations are:
$$Circle1: (x-(1+2\surd2))^2+(y-(2-2\surd2))^2=4^2$$
$$Circle2: (x-(1-2\surd2))^2+(y-(2+2\surd2))^2=4^2$$
A: Hint. The radius of each circle connecting its center to point $(1,2)$ is perpendicular to the tangent line that you found, so these radii have slope the negative reciprocal of $1$, that is $-1$. So the two centers belong to the line $(y-2)=-(x-1)$, and are distance $4$ from $(1,2)$, so you could find these centers, and from there the equations of the circles. 
A: suppose the center is $(a,b)$. as radius is $2$ the circle must be:
$$
(x-a)^2+(y-b)^2 = 4
$$
considering the slope of the tangent at (1,2) gives:
$$
(1-a) = -(2-b)
$$
and now also using the fact that $(1,2)$ is on the circle gives:
$$
2(1-a)^2 = 4 \\
a= 1 \pm \sqrt{2}
$$
and the corresponding values of $b$ are obtained from $b=3-a$
A: Or you can use geometry:
Since the tangent and normal line are perpendicular by definition, construct two right triangles from the point $(1,2)$, where the normal line is acting as the hypotenuse. Because the slope of the normal line has an absolute value of one, in essence you construct an isosceles right triangle $(45-45-90)$. Given that the hypotenuse is a length of $4$, that means each leg has a length of $2\sqrt{2}$. Just move accordingly along the normal line.
