Find the $x$ coordinate of point $C$

We want the area of the triangle to be equal to its semiperimeter, and so $\begin{aligned} 10 & = \; \tfrac12\left[4 + \sqrt{(x-2)^2 + 25} + \sqrt{(x+2)^2+25}\right] \\ 16 & = \; \sqrt{(x-2)^2 + 25} + \sqrt{(x+2)^2+25} \\ 256 & = \; 2x^2 + 58 + 2\sqrt{(x^2 + 29)^2 - 16x^2} \\ 99 - x^2 & = \; \sqrt{x^4 + 42x^2 + 841} \\ (99 - x^2)^2 & = \; x^4 + 42x^2 + 841 \\ 240x^2 & = \; 8950 \end{aligned}$ and hence $x^2 = \tfrac{112}{3}$ , so that $x = \tfrac43\sqrt{21} = \boxed{6.1101}$ .

I moved the x- and y-axis to the center of the circle and moved it back afterwards. Then I used polar lines with two parameters: the x-co-ordinate of the top of the triangle and one of the base points. The polar lines have the tangent points in common, I found them in terms of the two parameters. The points also lie on the circle. That determines the parameters and thus the position of the top.

I never heard about the theorem you used. Seems interesting though.

This seemed just too cute to leave on my screen.

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from sympy import Point, Triangle, solve, N
from sympy.abc import x

inradius = 1
A,B,C = map(Point, [(-2,0), (2,0), (x,5)])
T = Triangle(A, B, C)
eqn = 2*T.area/T.perimeter - inradius
soln = solve(eqn, x)[1]
print(soln, "≈", N(soln))

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4*sqrt(21)/3 ≈ 6.11010092660779