Solution:
\( (4,6), (-1,5), (0,0) \) and \( (k,3k) \) radius \( r \).
Let centre be \( (h,p) \)
\( O \) is circumcentre,

\( h^2 + p^2 = r^2 \)
\( (h+1)^2 + (p-5)^2 = r^2 \)
\( \Rightarrow h^2 + 1 + 2h + p^2 + 25 - 10p = r^2 \)
\( \Rightarrow h^2 + p + 2h - 10p + 26 = r^2 \)
\(\implies r^2 + 2h - 10p + 26 = r^2\)
\( \Rightarrow 2h - 10p = -26 \Rightarrow h - 5p = -13 \) \((1)\)
and
\( (h-4)^2 + (p-6)^2 = r^2 \)
\( \Rightarrow h^2+ 16 - 8h + p^2 + 36 - 12p = r^2 \)
\( \Rightarrow 8h + 12p = +52 \)
\( \Rightarrow 2h + 3p = 13 \) \((2)\)
\(\underline { 2h - 10p = -26} \)
\( 13p = 39 \Rightarrow p = 3 \)
\( \Rightarrow h = -13 + 5 \times 3 = -13 + 15 = 2 \)
\( \therefore \) centre \( (2,3) \) \( \therefore \) radius \( r = \sqrt{h^2 + p^2} = \sqrt{9 + 4} \Rightarrow r = \sqrt{13} \)
\( \therefore (k-2)^2 + (3k-3)^2 = r^2 = 13 \)
\( \Rightarrow k^2 + 4 - 4k + 9k^2 + 9 - 18k = 13 \)
\( \Rightarrow 10k^2 - 22k = 0 \Rightarrow 2k(5k-11) = 0 \)
\( \Rightarrow k = 0, \; \frac{11}{5} \)
\( \therefore k = \frac{11}{5} \)
\( \therefore \; 10k + 2r^2 = 10 \times \frac{11}{5} + r^2 = 22 + 13 = 35 \)

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