Increasing the radius of a cylinder by $ 6$ units increased the volume by $ y$ cubic units. Increasing the altitude of the cylinder by $ 6$ units also increases the volume by $ y$ cubic units. If the original altitude is $ 2$, then the original radius is:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 6\pi \qquad\textbf{(E)}\ 8$
1. Let the original radius of the cylinder be \( r \) and the original altitude (height) be \( h = 2 \).
2. The volume \( V \) of the cylinder is given by the formula:
   \[
   V = \pi r^2 h = \pi r^2 \cdot 2 = 2\pi r^2
   \]
3. When the radius is increased by 6 units, the new radius is \( r + 6 \). The new volume \( V_{\text{new}} \) is:
   \[
   V_{\text{new}} = \pi (r + 6)^2 \cdot 2
   \]
4. The increase in volume \( y \) is given by:
   \[
   y = V_{\text{new}} - V = \pi (r + 6)^2 \cdot 2 - 2\pi r^2
   \]
5. Simplify the expression for \( y \):
   \[
   y = 2\pi (r + 6)^2 - 2\pi r^2
   \]
   \[
   y = 2\pi \left( r^2 + 12r + 36 \right) - 2\pi r^2
   \]
   \[
   y = 2\pi r^2 + 24\pi r + 72\pi - 2\pi r^2
   \]
   \[
   y = 24\pi r + 72\pi
   \]
6. When the altitude is increased by 6 units, the new altitude is \( h + 6 = 2 + 6 = 8 \). The new volume \( V_{\text{new}} \) is:
   \[
   V_{\text{new}} = \pi r^2 \cdot 8
   \]
7. The increase in volume \( y \) is given by:
   \[
   y = V_{\text{new}} - V = \pi r^2 \cdot 8 - 2\pi r^2
   \]
8. Simplify the expression for \( y \):
   \[
   y = 8\pi r^2 - 2\pi r^2
   \]
   \[
   y = 6\pi r^2
   \]
9. Equate the two expressions for \( y \):
   \[
   24\pi r + 72\pi = 6\pi r^2
   \]
10. Divide both sides by \( 6\pi \):
    \[
    4r + 12 = r^2
    \]
11. Rearrange the equation to form a quadratic equation:
    \[
    r^2 - 4r - 12 = 0
    \]
12. Solve the quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -12 \):
    \[
    r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
    \]
    \[
    r = \frac{4 \pm \sqrt{16 + 48}}{2}
    \]
    \[
    r = \frac{4 \pm \sqrt{64}}{2}
    \]
    \[
    r = \frac{4 \pm 8}{2}
    \]
13. This gives two solutions:
    \[
    r = \frac{4 + 8}{2} = 6 \quad \text{and} \quad r = \frac{4 - 8}{2} = -2
    \]
14. Since the radius cannot be negative, we discard \( r = -2 \).

The final answer is \( \boxed{6} \).