A can contains a mixture of two liquids A and B in the ratio $7:5$. When $9$ litres of the mixture are drawn and replaced by the same amount of liquid $B$, the ratio of $A$ and $B$ becomes $7:9$. How many litres of liquid A was contained in the can initially?
$\textbf{(A)}~18$
$\textbf{(B)}~19$
$\textbf{(C)}~20$
$\textbf{(D)}~\text{None of the above}$
1. Let the initial quantities of liquids A and B be \(7x\) and \(5x\) liters respectively, where \(x\) is a common multiplier.
2. When 9 liters of the mixture are drawn, the amount of liquid A and B removed will be in the same ratio \(7:5\). Therefore, the amount of liquid A removed is \(\frac{7}{12} \times 9 = \frac{63}{12} = 5.25\) liters, and the amount of liquid B removed is \(\frac{5}{12} \times 9 = \frac{45}{12} = 3.75\) liters.
3. After removing 9 liters of the mixture, the remaining quantities of liquids A and B are:
   \[
   7x - 5.25 \quad \text{and} \quad 5x - 3.75
   \]
4. We then add 9 liters of liquid B to the mixture. The new quantities of liquids A and B are:
   \[
   7x - 5.25 \quad \text{and} \quad 5x - 3.75 + 9 = 5x + 5.25
   \]
5. According to the problem, the new ratio of liquids A and B is \(7:9\). Therefore, we set up the equation:
   \[
   \frac{7x - 5.25}{5x + 5.25} = \frac{7}{9}
   \]
6. Cross-multiplying to solve for \(x\):
   \[
   9(7x - 5.25) = 7(5x + 5.25)
   \]
   \[
   63x - 47.25 = 35x + 36.75
   \]
7. Simplifying the equation:
   \[
   63x - 35x = 36.75 + 47.25
   \]
   \[
   28x = 84
   \]
   \[
   x = 3
   \]
8. The initial quantity of liquid A is:
   \[
   7x = 7 \times 3 = 21 \text{ liters}
   \]

The final answer is \( \boxed{ 21 } \) liters.