Consider two parabolas
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Find the equation of circle of minimum radius touching both the parabolas.
$\displaystyle p_1: y-3=(x+1)^2$
$\displaystyle p_2: x-3=(y+1)^2$
$\displaystyle p_2$ is the image of $\displaystyle p_1$ of a reflection about the line with the equation y = x. Thus the circle in question touches the parabolae at those points where the gradient is +1.
I'll take $\displaystyle p_1$:
$\displaystyle y = x^2+2x+3~\implies~y'=2x+2$
$\displaystyle 2x+2=1~\implies~x=-\dfrac12$
Therefore the tangentpoint is $\displaystyle T\left(-\dfrac12\ ,\ \dfrac{13}4\right)$
A line perpendicular to the angle bisector of the first quadrant (do say first median?) passing through T crosses this angle bisector in the center of the circle in question:
$\displaystyle y-\dfrac{13}4 = -1\left(x+\dfrac12\right)~\implies~y=-x+\dfrac{11}4$ intersects y = x at
$\displaystyle A\left(\dfrac{11}8\ ,\ \dfrac{11}8\right)$
The radius is the segment $\displaystyle r = |\overline{TA}|$
1. The question asks you to find the smallest circle which is tangent to the two parabolae. I've drawn some of those tangent circles (see attachment). You'll get the smallest circle if the tangent points are the endpoints of a diameter, that means the radius is the shortest distance between one parabola and the angle bisector of the first quadrant and the diameter must be perpendicular to the angle bisector.
2. If a circle and a parabola are tangent to each other then they have a common tangent. This tangent must be perpendicular to the radius of the circle. The radius is perpendicular to the angle bisector and thus the tangent must be parallel to the angle bisector. That means the slope of the tangent is equal to the slope of the parabola (in the tangent point) and it is equal to the slope of th angle bisector.
3. Since I knew the slope of the angle bisector I used this fact to calculate the coordinates of the tangent point at the parabola.