Question:

Step 1: Determine the radii of the circles and relevant segment lengths. Given that $DC$ is the diameter of the large circle centered at $A$, and $DC = 12$ units. The radius of the large circle, $R$, is $DC/2 = 12/2 = 6$ units. Therefore, $AD = AC = R = 6$ units. Given that $AC$ is the diameter of the smaller circle centered at $B$. The radius of the smaller circle, $r_{small}$, is $AC/2 = 6/2 = 3$ units. Therefore, $AB = BC = r_{small} = 3$ units. Since $A$ is the center of $DC$ and $B$ is the center of $AC$, the points $D, A, B, C$ are collinear in that order. We can find the length of $BD$: $BD = AD + AB = 6 + 3 = 9$ units. Step 2: Calculate $\sin\theta$. Let $\theta = \angle EDC$. The line segment $DE$ is tangent to the smaller circle at point $F$. When a line is tangent to a circle, the radius drawn to the point of tangency is perpendicular to the tangent line. Thus, $BF \perp DE$, making $\triangle BFD$ a right-angled triangle with the right angle at $F$. In $\triangle BFD$, $BF$ is the radius of the smaller circle, so $BF = r_{small} = 3$ units. The hypotenuse of $\triangle BFD$ is $BD = 9$ units (calculated in Step 1). Using the definition of sine in $\triangle BFD$: $$ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BF}{BD} $$ $$ \sin\theta = \frac{3}{9} = \frac{1}{3} $$ Step 3: Calculate $\cos\theta$. Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $$ \cos^2\theta = 1 - \sin^2\theta $$ Substitute the value of $\sin\theta = 1/3$: $$ \cos^2\theta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} $$ Since $\theta$ is an angle in a right-angled triangle ($\triangle BFD$), it must be acute, so $\cos\theta > 0$. $$ \cos\theta = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3} $$ Step 4: Calculate the length of $DE$. We assume that point $E$ lies on the large circle. This is a common geometric interpretation in such problems when a chord length is requested. Since $DC$ is the diameter of the large circle and $E$ is a point on its circumference, $\triangle DEC$ is a right-angled triangle with $\angle DEC = 90^\circ$. In $\triangle DEC$, we can use the definition of cosine for $\angle EDC = \theta$: $$ \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{DE}{DC} $$ Rearranging to solve for $DE$: $$ DE = DC \cos\theta $$ Substitute the values $DC = 12$ and $\cos\theta = \frac{2\sqrt{2}}{3}$: $$ DE = 12 \times \frac{2\sqrt{2}}{3} $$ $$ DE = 4 \times 2\sqrt{2} $$ $$ DE = 8\sqrt{2} $$ The final answer is $\boxed{8\sqrt{2}}$.