Equation of Incircle in Equilateral Triangle

To find the equation of the circle $C$, we need to determine its centre and radius.

1. Given Information:

• The circle $C$ has radius $r = 1$.
• The line $PQ$ has the equation:
  $L(x, y) = \sqrt{3}x + y - 6 = 0$
• The point of contact of $C$ with $PQ$ is:
  $D = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$
• The origin $(0, 0)$ and the centre of $C$ lie on the same side of the line $PQ$.

2. Finding the Centre of Circle $C$:
Since $PQ$ is tangent to the circle at point $D$, the line joining the centre $(h, k)$ to $D$ is perpendicular to $PQ$.

The line $PQ$ has a normal vector:
$\vec{n} = (\sqrt{3}, 1)$
The unit normal vector is:
$\hat{n} = \frac{(\sqrt{3}, 1)}{\sqrt{(\sqrt{3})^2 + 1^2}} = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

Since the centre lies at a distance of $r = 1$ along the normal to the tangent at $D$, the coordinates of the centre are given by:
$(h, k) = D \pm r\hat{n} = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \pm 1 \cdot \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

This gives two possible centres:

1. $(h_1, k_1) = \left(\frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, \frac{3}{2} - \frac{1}{2}\right) = (\sqrt{3}, 1)$
2. $(h_2, k_2) = \left(\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2}, \frac{3}{2} + \frac{1}{2}\right) = (2\sqrt{3}, 2)$

3. Applying the Position of Centre Relative to Line $PQ$:
The value of the line equation at the origin $(0, 0)$ is:
$L(0, 0) = \sqrt{3}(0) + 0 - 6 = -6 < 0$

For the centre to be on the same side of the line as the origin, $L(h, k)$ must also be negative:

• Testing $(h_1, k_1) = (\sqrt{3}, 1)$:
  $L(\sqrt{3}, 1) = \sqrt{3}(\sqrt{3}) + 1 - 6 = 3 + 1 - 6 = -2 < 0$
• Testing $(h_2, k_2) = (2\sqrt{3}, 2)$:
  $L(2\sqrt{3}, 2) = \sqrt{3}(2\sqrt{3}) + 2 - 6 = 6 + 2 - 6 = 2 > 0$

Thus, the centre of the circle is:
$(h, k) = (\sqrt{3}, 1)$

4. Equation of Circle $C$:
With centre $(\sqrt{3}, 1)$ and radius $r = 1$, the equation of the circle is:
$(x - \sqrt{3})^2 + (y - 1)^2 = 1$

Therefore, the correct option is (D).