Let $A_1A_2A_3$ be the triangle and $(I_i,k_i)$ be the circle next to the vertex $A_i,$ where $i=1,2,3.$ Let $I$ be the triangle’s incenter. Let $x_2$ and $x_3$ be the projection of $II_2$ and $II_3$ on $A_2A_3.$ Then we obviously have the following: $x_2^2=(k+k_2)^2-(k-k_2)^2=4kk_2,\ \ (1)$ and similarly $x_3^2=4kk_3.\ \ (2)$ Also, from triangle $II_2I_3$ we have: $(I_2I_3)^2 =(x_2+x_3)^2+(k_3-k_2)^2=(II_2)^2+(II_3)^2+2\cdot II_2 \cdot II_3\cdot\sin{\frac{A_1}{2}}.\ \ (3)$ From the above three expressions we get: $k(k_2+k_3)-k_2k_3+4k\sqrt{k_2k_3}-k^2= (k+k_2)(k+k_3)\sin{\frac{A_1}{2}}.\ \ (4)$ However, $\sin{\frac{A_1}{2}}=\frac{k-k_1}{k+k_1},$ and summing equation $(4)$ over $i=1,2,3$ we finally get: $2k(k_1+k_2+k_3)-(k_1k_2+k_2k_3+k_1k_3)+4k(\sqrt{k_1k_2}+\sqrt{k_2k_3}+\sqrt{k_1k_3})-3k^2=(k+1)(k+4)\frac{k-9}{k+9}+(k+4)(k+9)\frac{k-1}{k+1}+(k+1)(k+9)\frac{k-4}{k+4}.$ Substituting for $k_1=1,$ $k_2=4,$ and $k_3=9,$ and rearranging, we get $k(k-11)(3k^3+39k^2+121k+37)=0.$ Because $k>9,$ we have $k=11.$