Finding minimum value of x such that GCD(A+x,B+x) = C where A , B ,C are given I need to add Minimum non-negative Integer such that I can get the desired GCD(a+x,b+x)
Let say A=12 & B=26 
For GCD(12+x,26+x) = 1 , x should be 1 
For GCD(12+x,26+x) = 2 , x should be 0 
For GCD(12+x,26+x) = 7 , x should be 9 
For GCD(12+x,26+x) = 14 , x should be 2 

Your question title says you want to find the minimum value of a non-negative integer $x$ such that $\gcd(A + x, B + x) = C$, where $A, B, C$ are given. First, note there's no such integer in certain cases. In particular, since $C$ divides $A + x$ and $B + x$, it must also divide their difference, i.e., $C \mid (A + x) - (B + x) = A - B$. Thus, a minimum requirement is that
$$A \equiv B \pmod C \tag{1}\label{eq1}$$
Next, consider the special case of $A = B$. If $A \le -C$, then $x = -C - A$ works. Otherwise, if $A \le C$, then $x = C - A$ works. Finally, if $A \gt C$, then the gcd would always be $A + x \gt C$, so there are no solutions.
If \eqref{eq1} is satisfied and $A \neq B$, next note $A$ and $B$ can be rewritten as
$$A = mC + r, B = nC + r, \; \text{ where } \; m,n,r \in \mathbb{Z}, \; m \neq n, \; 0 \le r \lt C \tag{2}\label{eq2}$$
Thus, $A + x = mC + (r + x)$, so $C \mid A + x$ requires $C \mid r + x$. In general, this requires
$$x = kC - r, \; k \ge 0 \tag{3}\label{eq3}$$
The smallest non-negative value of $x$ which might work is $0$ if $r = 0$, else it's $C - r$. For $B + x = nC + (r + x)$, it's also true that $C \mid B + x$ in any general case.
The one thing you need to be careful of is that there are no other common factors so the GCD would then be a multiple of $C$. In particular, \eqref{eq3}, gives $A + x = C(m + k)$ and $B + x = C(n + k)$. Thus, you want to find the smallest non-negative integer $k$ such that $x$ from \eqref{eq3} is non-negative and $\gcd(m + k, n + k) = 1$. Note there'll always be such a $k$. To see this, the gcd of $m + k$ and $n + k$ must divide the difference, i.e., $m - n$. Since $m \neq n$, then $\left|m - n\right|$ is either $1$, in which case the gcd is always $1$, or it's a product of $1$ or more primes, i.e., $\left|m - n\right| = \prod_{i = 1}^j p_i^{e_i}$ for some $j \ge 1$, primes $p_i$ and integers $e_i \ge 1$. In the latter case, for each $i$, note that $m \equiv n \equiv r_i \pmod p_i$ for some $0 \le r_i \le p_i - 1$. Since all primes are $\ge 2$, there are always at least $2$ possible values for each $r_i$. For each one, choose any value such its sum with $r_i$ is not congruent to $0$. No $p_i$ would divides $n + k$ or $m + k$. Thus, any prime factors of $\gcd(n + k, m + k)$ must be different than any $p_i$. However, since any prime factors of the gcd must divide the difference, this means there are no prime factors of the gcd, i.e., $\gcd(n + k, m + k) = 1$. For each possible combination of these congruences, the method described to handle more than $2$ numbers section in Extended Euclidean algorithm shows  you can always find non-negative integers $k$ satisfying those congruences. Thus, choose the smallest non-negative integer $k$ among all those values so that $x$ given in \eqref{eq3} is non-negative.
To demonstrate how this works, here's how to apply this to your examples with $A = 12$ and $B = 26$. First, as $26 - 12 = 14 = 2 \times 7$, the only possible values of $C$ are $1, 2, 7, 14$, as you've shown. For $C = 1$, since $\gcd(12, 26) = 2$, as $12 = 12 \times 1 + 0$ and $26 = 26 \times 1 + 0$, but $\gcd(12 + 1, 26 + 1) = 1$, you need to use $x = 1$. For $C = 2$, just $x = 0$ works. For $C = 7$, since $12 = 1 \times 7 + 5$, $26 = 3 \times 7 + 5$, and $\gcd(1 + 1, 3 + 1) = 2$, you can't use $x = 7 - 5 = 2$. Instead, since $\gcd(1 + 2, 3 + 2) = 1$, you need to use $x = 14 - 5 = 9$ instead. Finally, for $C = 14$, since $12 = 0 \times 14 + 12$, $26 = 1 \times 14 + 12$ and $\gcd(0 + 1, 1 + 1) = 1$, you can just use $x = 14 - 12 = 2$.
