The length $l$ of a tangent, drawn from a point $A$ to a circle, is $\frac43$ of the radius $r$.  The (shortest) distance from $A$ to the circle is:
$ \textbf{(A)}\ \frac{1}{2}r \qquad\textbf{(B)}\ r\qquad\textbf{(C)}\ \frac{1}{2}l\qquad\textbf{(D)}\ \frac23l \qquad\textbf{(E)}\ \text{a value between r and l.} $
1. Let \( O \) be the center of the circle and \( A \) be the point from which the tangent is drawn. Let \( l \) be the length of the tangent from \( A \) to the circle, and \( r \) be the radius of the circle. According to the problem, \( l = \frac{4}{3}r \).

2. By the Pythagorean theorem, in the right triangle \( \triangle OAT \) (where \( T \) is the point of tangency), we have:
   \[
   AO^2 = OT^2 + AT^2
   \]
   Here, \( OT = r \) and \( AT = l \). Substituting the given values, we get:
   \[
   AO^2 = r^2 + \left(\frac{4}{3}r\right)^2
   \]

3. Simplify the equation:
   \[
   AO^2 = r^2 + \left(\frac{16}{9}r^2\right)
   \]
   \[
   AO^2 = r^2 + \frac{16}{9}r^2
   \]
   \[
   AO^2 = \frac{9}{9}r^2 + \frac{16}{9}r^2
   \]
   \[
   AO^2 = \frac{25}{9}r^2
   \]

4. Taking the square root of both sides, we get:
   \[
   AO = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r
   \]

5. The shortest distance from point \( A \) to the circle is the distance from \( A \) to the center \( O \) minus the radius \( r \):
   \[
   \text{Distance} = AO - r = \frac{5}{3}r - r = \frac{5}{3}r - \frac{3}{3}r = \frac{2}{3}r
   \]

Thus, the shortest distance from \( A \) to the circle is \( \frac{2}{3}r \).

The final answer is \( \boxed{\frac{2}{3}r} \)