Solve following system equations:
\[\left\{ \begin{array}{c}
3x+4y=26\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \\ 
\sqrt{x^2+y^2-4x+2y+5}+\sqrt{x^2+y^2-20x-10y+125}=10\  \end{array}
\right.\ \ \] 



1. Start with the given system of equations:
   \[
   \left\{ 
   \begin{array}{c}
   3x + 4y = 26 \\
   \sqrt{x^2 + y^2 - 4x + 2y + 5} + \sqrt{x^2 + y^2 - 20x - 10y + 125} = 10 
   \end{array}
   \right.
   \]

2. Solve the first equation for \( y \):
   \[
   3x + 4y = 26 \implies 4y = 26 - 3x \implies y = \frac{26 - 3x}{4}
   \]

3. Substitute \( y = \frac{26 - 3x}{4} \) into the second equation:
   \[
   \sqrt{x^2 + \left(\frac{26 - 3x}{4}\right)^2 - 4x + 2\left(\frac{26 - 3x}{4}\right) + 5} + \sqrt{x^2 + \left(\frac{26 - 3x}{4}\right)^2 - 20x - 10\left(\frac{26 - 3x}{4}\right) + 125} = 10
   \]

4. Simplify the expressions inside the square roots. Let’s denote:
   \[
   A(x, y) = (x, y), \quad B = (2, -1), \quad C = (10, 5)
   \]
   The second equation can be interpreted geometrically as:
   \[
   \sqrt{(x - 2)^2 + (y + 1)^2} + \sqrt{(x - 10)^2 + (y - 5)^2} = 10
   \]
   This represents the sum of distances from point \( A \) to points \( B \) and \( C \).

5. According to the triangle inequality, the sum of the distances from a point to two other points is equal to the distance between the two points if and only if the point lies on the line segment joining the two points. Therefore, point \( A \) must lie on the line segment \( BC \).

6. Find the equation of the line passing through points \( B \) and \( C \):
   \[
   \text{slope} = \frac{5 - (-1)}{10 - 2} = \frac{6}{8} = \frac{3}{4}
   \]
   The equation of the line is:
   \[
   y - (-1) = \frac{3}{4}(x - 2) \implies y + 1 = \frac{3}{4}x - \frac{3}{2} \implies y = \frac{3}{4}x - \frac{5}{2}
   \]

7. Since \( A \) lies on this line, substitute \( y = \frac{26 - 3x}{4} \) into the line equation:
   \[
   \frac{26 - 3x}{4} = \frac{3}{4}x - \frac{5}{2}
   \]

8. Clear the fractions by multiplying through by 4:
   \[
   26 - 3x = 3x - 10
   \]

9. Solve for \( x \):
   \[
   26 + 10 = 3x + 3x \implies 36 = 6x \implies x = 6
   \]

10. Substitute \( x = 6 \) back into \( y = \frac{26 - 3x}{4} \):
    \[
    y = \frac{26 - 18}{4} = \frac{8}{4} = 2
    \]

11. Therefore, the solution to the system of equations is:
    \[
    (x, y) = (6, 2)
    \]

The final answer is \(\boxed{(x, y) = (6, 2)}\)