Lopsidedness

Clearly $B$ is an unit circle with center at origin.

Since every chord is orthogonal to the radius, $A$ must pass through point $(0,p)$ and $C$ must pass through point $(0,-p)$ .
Now, we can clearly see that $A$ and $C$ are reflections of each other (by $x$ -axis). Now, $B$ is divided into $3$ regions.

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We notice that lopsidedness is minimized when all regions have an equal area.
Such $p$ exist because when $p$ goes from $0$ to $\infty$ , the combined area of the segments of $A$ and $C$ which form one of the regions goes from the area of $B$ to $0$ .

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Now since we want to have equal area in all regions, we must have $\text{area of B} = 3 \times \text{combined area of the segments}$ .

Area of $B$ is clearly equal to $\pi$ and $\text{area of the segment} = \text{area of the sector} - \text{area of the isosceles triangle in the sector}$

Area of the triangle equals $\dfrac{2 \times p}{2} = p$ (Clearly $p$ is the height of the isosceles triangle and $2$ is the base since it is the chord) and
the area of the sector equals $\dfrac{2 \times \arctan\left(\dfrac1p\right)}{2\pi} \times \pi \times \sqrt{p^2 + 1 }^2 = \arctan\left(\dfrac1p\right) \times \left(p^2 + 1\right)$ .
( $\tan{\alpha} = \dfrac{\text{opposite}}{\text{adjacent}}$ in a right triangle, isosceles triangle can be divided into $2$ identical right triangles)

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Thus, the area of the segments combined is $2 \times \left(\arctan\left(\dfrac1p\right) \times \left(p^2 + 1\right) - p\right) = \dfrac\pi3$ which implies that $\left(p^2 + 1\right) \arctan\left(\dfrac1p\right) = p + \dfrac\pi6$ .

Hence, $K = \boxed{6}$ .

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If someone wants to show me a good tool for making and image to clarify this solution, please leave a comment.

It is obvious that circle B is the unit circle, with area $\pi$ . Circles A and C are bigger and lie symmetrically around the origin:

For sufficiently small $p$ values, the yellow region will be largest, but eventually the other two regions will be bigger. Thus the desired minimum is reached when the three areas are equally large, i.e. the area of the yellow region is $\pi/3$ .

We focus on the top half of this region, which should have area $\pi/6$ . The three given points for circle C lie on the three vertices of a rectangle; therefore the circle also goes through its fourth vertex, and the center of the circle coincides with the center of the rectangle; it is the point $(0, -p)$ . Applying Pythagoras shows that the radius of circle C is $R_C = \sqrt{1 + p^2}$ .

The yellow area is found by subtracting the area of a circle sector (the blue lines) minus the area of a triangle.

The area of the triangle is $\tfrac12\cdot 2 \cdot p = p$ .

The area of the circle sector is $\pi R_C^2 \cdot 2\theta/360^\circ$ , and we see readily that $\tan\theta = 1/p$ . Thus

$\text{yellow area} = \pi R_C^2 \cdot \frac{2\theta}{2\pi} - p = (1 + p^2)\left(\arctan \frac 1p\right) - p;$

since the yellow area is $\pi/6$ , this becomes

$(1 + p^2)\left(\arctan \frac 1p\right) = p + \frac{\pi}6,$ showing that $\boxed{K = 6}$ .