Identifying a circle

Let ${{C}_{1}}$ be the circle in question. We can locate the center $C$ of ${{C}_{1}}$ as the intersection of two loci:

• the perpendicular bisector $l$ of $\overline{AB}$ , since $C$ is equidistant from $A$ and $B$ and

• the parabola ${{C}_{2}}$ with focus $A$ and directrix the x-axis, since $C$ is equidistant from $A$ and the x-axis.

To find the equation of $l$ , $\begin{aligned} M\left( x,y \right)\in l & \Leftrightarrow \left| \overline{AM} \right|=\left| \overline{BM} \right|\Leftrightarrow {{\left| \overline{AM} \right|}^{2}}={{\left| \overline{BM} \right|}^{2}} \\ & \Leftrightarrow {{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( x-5 \right)}^{2}}+{{\left( y-7 \right)}^{2}} \\ & \Leftrightarrow 14x+8y-61=0 \ \ \ \ \ (1)\\ \end{aligned}$ The semi-latus rectum of the parabola is $p=d\left( A,{x}'x \right)=3$ , hence the equation of the parabola is $y=\dfrac{1}{2p}{{\left( x-{{x}_{A}} \right)}^{2}}+\dfrac{p}{2}\Leftrightarrow y=\dfrac{1}{6}{{\left( x+2 \right)}^{2}}+\frac{3}{2} \ \ \ \ \ (2)$ Solving the system of $(1)$ and $(2)$ we get $x=\dfrac{-29\pm \sqrt{1365}}{4},\ \ \ \ \ \ y=\dfrac{325\mp \sqrt{1365}}{16}$ The y-coordinate of $C$ corresponds to the radius of the circle as well, thus the smallest circle has radius $R=\dfrac{325-\sqrt{1365}}{16}\approx 4.14867$ For the answer, $\left\lfloor {{10}^{4}}R \right\rfloor =\boxed{41486}$ .

Let the center of the circle be $C(x, y)$ . Since the circle is tangent to the $x$ -axis, its radius is $y$ .

Then by the distance equation, $AC^2 = (x + 2)^2 + (y - 3)^2 = y^2$ and $BC^2 = (x - 5)^2 + (y - 7)^2 = y^2$ .

These two equations solve to $x = \cfrac{-29 \pm \sqrt{1365}}{4}$ and $y = \cfrac{325 \mp 7\sqrt{1365}}{16}$ , the smaller radius being $R = y = \cfrac{325 - 7\sqrt{1365}}{16} \approx 4.14867$ .

Therefore, $\lfloor 10^4 R \rfloor = \boxed{41486}$ .