Find radius of circle which is tangential to parabola. 
Given $r < 57$, let $C$ denote a circle centered at ($57$, $r$) with
  radius $r$. 
$C$ is tangent to a parabola, $y=x^2+r$, from the outside in the first
  quadrant.
Find the value of $r$.

I am not sure how to approach the above problem. I started with implicit differentiation of the general circle formula.


*

*$r^2 = (x-57)^2+(y - r)^2$

*$D(r^2) = D\left((x-57)^2+( y - r)\right)$

*$0 = 2x - 114 + y'(2y-2r)$

*$y' = 114-\dfrac{2x}{2y-2r}$
For parabola:


*

*$D(y)=D(x^2+r)$

*$y'=2x$
Which gives me the formula of the gradient of the tangent to the circle. However, I am unsure how to proceed from here. What does the question means by from the outside in the first quadrant?
I am actually modeled the parabola and circle in an online graph but I fail to see how does the first quadrant requirement comes into play since $r<57$. 

Let point of intersect be $(h,k)$. We get two equations, one from equating slopes and other from equating $y$ coordinate. $(h,k)$ satisfies both equations.
$$\frac{114-2h}{2k-2r} = 2h \tag{1}$$
$$(h-57)^2 + (k-r)^2 = r^2 \tag{2}$$
From first equation we can write:
$$114-2h = 4h^3$$
Since this is a cubic representing $x$ coordinate of point of intersection, we need only one real root. Which is the case here because  derivative of cubic $4h^3+2h-114$ is positive for all real. The exact value might be cumbersome to calculate.
Now that we have $h$, use equation two to get $r$:
$$(h-57)^2 +h^4 = r^2$$
