Step-by-step Derivation

1. Identify Given Information

   • The circle C has a radius r = 1.
   • The equation of the line PQ, which is tangent to the circle, is $L: \sqrt 3 x + y - 6 = 0$.
   • The point of contact of the circle with the line PQ is $D \left( {{{3\,\sqrt 3 } \over 2}, \,{3 \over 2}} \right)$.
   • The origin (0, 0) and the center of the circle C, let's call it O'(h, k), are on the same side of the line PQ.

2. Goal: Find the Equation of the Circle The equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$. Since we know r = 1, we need to find the coordinates of the center O'(h, k).

3. Using Properties of Tangents The radius of a circle is perpendicular to the tangent at the point of contact. This means the line segment O'D is perpendicular to the line PQ.

4. Find the Slope of the Normal

   • First, find the slope of the line PQ ($\sqrt 3 x + y - 6 = 0$). Rearranging the equation to the slope-intercept form y = mx + c gives $y = -\sqrt 3 x + 6$. So, the slope of PQ is $m_{PQ} = -\sqrt 3$.
   • Since O'D is perpendicular to PQ, the product of their slopes is -1. $m_{O'D} \times m_{PQ} = -1$ $m_{O'D} = -1 / m_{PQ} = -1 / (-\sqrt 3) = 1/\sqrt 3$.

5. Relate the Coordinates of the Center (h, k)

   • The slope of the line segment O'D can also be expressed using the coordinates of O'(h, k) and $D(3\sqrt{3}/2, 3/2)$: $m_{O'D} = {{k - 3/2} \over {h - 3\sqrt 3 / 2}}$
   • Equating the two expressions for $m_{O'D}$: ${{k - 3/2} \over {h - 3\sqrt 3 / 2}} = {1 \over {\sqrt 3 }}$ $\sqrt 3 (k - 3/2) = h - 3\sqrt 3 / 2$ $\sqrt 3 k - {3\sqrt 3 \over 2} = h - {3\sqrt 3 \over 2}$ $h = \sqrt 3 k$

6. Use the Radius Information

   • The distance between the center O'(h, k) and the point of contact D is equal to the radius r = 1.
   • Using the distance formula, $(O'D)^2 = r^2$: ${\left( {h - {{3\sqrt 3 } \over 2}} \right)^2} + {\left( {k - {3 \over 2}} \right)^2} = 1^2$
   • Substitute $h = \sqrt 3 k$ into this equation: ${\left( {\sqrt 3 k - {{3\sqrt 3 } \over 2}} \right)^2} + {\left( {k - {3 \over 2}} \right)^2} = 1$ ${\left[ {\sqrt 3 \left( {k - {3 \over 2}} \right)} \right]^2} + {\left( {k - {3 \over 2}} \right)^2} = 1$ $3{\left( {k - {3 \over 2}} \right)^2} + {\left( {k - {3 \over 2}} \right)^2} = 1$ $4{\left( {k - {3 \over 2}} \right)^2} = 1$ ${\left( {k - {3 \over 2}} \right)^2} = {1 \over 4}$ $k - {3 \over 2} = \pm {1 \over 2}$

7. Find Possible Centers

   • Case 1: $k - 3/2 = 1/2 \implies k = 3/2 + 1/2 = 2$. Then $h = \sqrt 3 k = 2\sqrt 3$. The center is $(2\sqrt 3, 2)$.
   • Case 2: $k - 3/2 = -1/2 \implies k = 3/2 - 1/2 = 1$. Then $h = \sqrt 3 k = \sqrt 3$. The center is $(\sqrt 3, 1)$.

8. Apply the Position Condition

   • We are given that the origin (0, 0) and the center O'(h, k) lie on the same side of the line $L: \sqrt 3 x + y - 6 = 0$. This means that L(0, 0) and L(h, k) must have the same sign.
   • Evaluate L at the origin: $L(0, 0) = \sqrt 3 (0) + (0) - 6 = -6$, which is negative.
   • Now, we check the two possible centers:
     • For $O'(2\sqrt 3, 2)$: $L(2\sqrt 3, 2) = \sqrt 3 (2\sqrt 3) + 2 - 6 = 6 + 2 - 6 = 2$, which is positive. This point is on the opposite side of the line from the origin.
     • For $O'(\sqrt 3, 1)$: $L(\sqrt 3, 1) = \sqrt 3 (\sqrt 3) + 1 - 6 = 3 + 1 - 6 = -2$, which is negative. This point is on the same side of the line as the origin.
   • Therefore, the correct center of the circle is $(\sqrt 3, 1)$.

9. Write the Final Equation

   • The center is $(h, k) = (\sqrt 3, 1)$ and the radius is r = 1.
   • The equation of the circle C is: $(x - \sqrt 3)^2 + (y - 1)^2 = 1^2$
   • This simplifies to $(x - \sqrt 3)^2 + (y - 1)^2 = 1$.

10. Compare with Options The derived equation $(x - \sqrt 3)^2 + (y - 1)^2 = 1$ matches option D.

Conclusion

The equation of circle C is ${\left( {x - \sqrt 3 } \right)^2} + {(y - 1)^2} = 1$.