The important information you'll find in the graphics.

The center of the Lester circle is Kimberling center X1116.

This center lies on the line X140-X523.

X140 is the midpoint of line segment ON.

But X523 is a center which lies on the line at infinity.

Therefore I think, that the center of the circle through O, N and K

is not yet contained in the 'Encylopedia of Triangle Centers'.

The link to this encyclopedia, where you can get more information:

[here]

The radius of the circle, on which O, N and K lie

can be calculated

with the help of the following formulas:

If you are interested in a calculation of this radius,

you can try to copy the following formulas

(e.g. into Mathematica).

R=     a*b*c/4/delta

delta=     1/4*((2*(a*a*b*b+b*b*c*c+c*c*a*a)-((a^4)+(b^4)+(c^4)))^(1/2))

ON=      1/2*(9*((a*b*c/((2*(a*a*b*b+b*b*c*c+c*c*a*a)-((a^4)+(b^4)+(c^4)))^(1/2)))^2)-(a^2+b^2+c^2))^(1/2)

OK=      (a*b*c)*((((a^4)+(b^4)+(c^4))-(a*a*b*b+b*b*c*c+c*c*a*a))^(1/2))/((((2*(a*a*b*b+b*b*c*c+c*c*a*a)-((a^4)+(b^4)+(c^4))))^(1/2))*(a^2+b^2+c^2)/2)

HK=      (4*((a*b*c)/((2*(a*a*b*b+b*b*c*c+c*c*a*a)-((a^4)+(b^4)+(c^4)))^(1/2)))^2-((a^4)+(b^4)+(c^4))/(a^2+b^2+c^2)-3*((a*b*c)*(a*b*c))/((a^2+b^2+c^2)^2))^(1/2)

NK=      (1/(2^(1/2))*((HK*HK+OK*OK-2*ON*ON)^(1/2)))

RONK=    ON*OK*NK/((2*(ON*ON*OK*OK+OK*OK*NK*NK+NK*NK*ON*ON)-(ON^4+OK^4+NK^4))^(1/2))

Postscript from Nov. 8, 2024:

The proof: Center X115 lies on the circle through O, N and K:

Contact

... to the author is maybe possible via e-mail under:   §@gmx.de   where   § = heisss

Subject: "Circle-through-O-N-K"