The correct answer is A).\n\nLet the given side lengths be $c = AB = 6$, $b = AC = 10$, and $a = BC = 14$. $AD$ is the angle bisector of $\angle A$.\n\n**Step 1: Apply the Angle Bisector Theorem.**\nAccording to the Angle Bisector Theorem, the bisector of an angle in a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. Thus, for $\triangle ABC$ and angle bisector $AD$:\n\n$\\frac{BD}{DC} = \\frac{AB}{AC} = \\frac{c}{b} = \\frac{6}{10} = \\frac{3}{5}$\n\nWe know that $BD + DC = BC = a = 14$. Let $BD = 3k$ and $DC = 5k$. Then $3k + 5k = 8k = 14$.\n\nSo, $k = \\frac{14}{8} = \\frac{7}{4}$.\n\nTherefore, $BD = 3 \\times \\frac{7}{4} = \\frac{21}{4}$ units and $DC = 5 \\times \\frac{7}{4} = \\frac{35}{4}$ units.\n\n**Step 2: Determine the radius of the circle.**\nThe circle is centered at $D$ and is tangent to side $AC$. The radius of the circle, let's call it $r$, is the perpendicular distance from the center $D$ to the line $AC$. Let $E$ be the foot of the perpendicular from $D$ to $AC$. Then $DE = r$.\n\nIn right-angled $\triangle DEC$, we have $DE = DC \\sin(\\angle C)$.\nTo find $r$, we need the value of $\\sin(\\angle C)$.\n\n**Step 3: Use the Cosine Rule to find $\\cos(\\angle C)$.**\nIn $\triangle ABC$, we can use the Cosine Rule to find $\\cos(\\angle C)$:\n\n$AB^2 = AC^2 + BC^2 - 2 \\times AC \\times BC \\cos(\\angle C)$\n$c^2 = b^2 + a^2 - 2ba \\cos(\\angle C)$\n\nSubstituting the given values:\n$6^2 = 10^2 + 14^2 - 2 \\times 10 \\times 14 \\cos(\\angle C)$\n$36 = 100 + 196 - 280 \\cos(\\angle C)$\n$36 = 296 - 280 \\cos(\\angle C)$\n$280 \\cos(\\angle C) = 296 - 36$\n$280 \\cos(\\angle C) = 260$\n$\\cos(\\angle C) = \\frac{260}{280} = \\frac{26}{28} = \\frac{13}{14}$\n\n**Step 4: Calculate $\\sin(\\angle C)$.**\nUsing the trigonometric identity $\\sin^2 \\theta + \\cos^2 \\theta = 1$:\n\n$\\sin^2(\\angle C) = 1 - \\cos^2(\\angle C)$\n$\\sin^2(\\angle C) = 1 - \\left(\\frac{13}{14}\\right)^2$\n$\\sin^2(\\angle C) = 1 - \\frac{169}{196}$\n$\\sin^2(\\angle C) = \\frac{196 - 169}{196} = \\frac{27}{196}$\n\nSince $\\angle C$ is an angle in a triangle, $\\sin(\\angle C)$ must be positive.\n$\\sin(\\angle C) = \\sqrt{\\frac{27}{196}} = \\frac{\\sqrt{9 \\times 3}}{\\sqrt{196}} = \\frac{3\\sqrt{3}}{14}$\n\n**Step 5: Calculate the radius $r$.**\nNow substitute the value of $\\sin(\\angle C)$ into the expression for $r$ from Step 2:\n\n$r = DC \\sin(\\angle C)$\n$r = \\frac{35}{4} \\times \\frac{3\\sqrt{3}}{14}$\n$r = \\frac{35 \\times 3\\sqrt{3}}{4 \\times 14}$\n\nSimplify the expression:\n$r = \\frac{(5 \\times 7) \\times 3\\sqrt{3}}{4 \\times (2 \\times 7)}$\n$r = \\frac{5 \\times 3\\sqrt{3}}{4 \\times 2}$\n$r = \\frac{15\\sqrt{3}}{8}$\n\nThus, the radius of the circle is $\\frac{15\\sqrt{3}}{8}$ units.\n\nThe final answer is $\\boxed{\\text{A}}$.\n