How many integers greater than $10$ and less than $100$, written in base-$10$ notation, are increased by $9$ when their digits are reversed?

$\textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$
1. Let the two-digit number be represented as \(10x + y\), where \(x\) and \(y\) are the digits of the number, and \(x\) is the tens digit and \(y\) is the units digit. Given that the number is increased by 9 when its digits are reversed, we can write the equation:
   \[
   10x + y + 9 = 10y + x
   \]

2. Simplify the equation:
   \[
   10x + y + 9 = 10y + x
   \]
   \[
   10x + y + 9 - x = 10y
   \]
   \[
   9x + y + 9 = 10y
   \]
   \[
   9x + 9 = 9y
   \]
   \[
   x + 1 = y
   \]

3. From the equation \(x + 1 = y\), we see that \(y\) is one more than \(x\). Since \(x\) and \(y\) are digits, they must be integers between 0 and 9. However, since the number must be greater than 10 and less than 100, \(x\) must be between 1 and 9 (inclusive).

4. Therefore, \(y\) must be between 2 and 9 (inclusive) because \(y = x + 1\).

5. The possible values for \(y\) are \(2, 3, 4, 5, 6, 7, 8, 9\). Each of these corresponds to a unique \(x\) value:
   - If \(y = 2\), then \(x = 1\)
   - If \(y = 3\), then \(x = 2\)
   - If \(y = 4\), then \(x = 3\)
   - If \(y = 5\), then \(x = 4\)
   - If \(y = 6\), then \(x = 5\)
   - If \(y = 7\), then \(x = 6\)
   - If \(y = 8\), then \(x = 7\)
   - If \(y = 9\), then \(x = 8\)

6. There are 8 such pairs \((x, y)\) that satisfy the condition \(y = x + 1\).

The final answer is \(\boxed{8}\).