The equation of circle C is: | vedclass

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(D) The centre of circle $C$ lies on the line passing through $D$ perpendicular to $PQ$. Thus,the centre of $C$ lies on $y - \frac{3}{2} = \left(\frac

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$(x - \sqrt{3})^2 + (y - 1)^2 = 1$

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(D) The centre of circle $C$ lies on the line passing through $D$ perpendicular to $PQ$. Thus,the centre of $C$ lies on $y - \frac{3}{2} = \left(\frac{1}{\sqrt{3}}\right)(x - \frac{3\sqrt{3}}{2}) \Rightarrow x = \sqrt{3}y$.Let the centre of the circle $C$ be $(\sqrt{3}y_1, y_1)$. Then,$(\frac{3\sqrt{3}}{2} - \sqrt{3}y_1)^2 + (\frac{3}{2} - y_1)^2 = 1$$\Rightarrow 3(\frac{3}{2} - y_1)^2 + (\frac{3}{2} - y_1)^2 = 1$$\Rightarrow 4(\frac{3}{2} - y_1)^2 = 1$ $\Rightarrow \frac{3}{2} - y_1 = \pm \frac{1}{2}$ $\Rightarrow y_1 = 1$ or $y_1 = 2$.Thus,the centre of $C$ can be $(\sqrt{3}, 1)$ or $(2\sqrt{3}, 2)$. Since the centre of the circle and the origin lie on the same side of $\sqrt{3}x + y - 6 = 0$,and $\sqrt{3}(0) + 0 - 6 Hence,the equation of the circle is $(x - \sqrt{3})^2 + (y - 1)^2 = 1$.

The equation of circle $C$ is:

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