Question

The equation of circle C is

• A. (x - 2$\sqrt{3}$)$^{2}$ + (y - 1)$^{2}$ = 1
• B. (x - 2$\sqrt{3}$)$^{2}$ + (y + $\frac{1}{2}$)$^{2}$ = 1
• C. (x - $\sqrt{3}$)$^{2}$ + (y + 1)$^{2}$ = 1
• D. (x - $\sqrt{3}$)$^{2}$ + (y - 1)$^{2}$ = 1

Answer 1

Answer: (D)

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

Answer 2

Let the center of circle C be $(h, k)$ and its radius be $r=1$. The distance from $(h, k)$ to the line $\sqrt{3}x + y - 6 = 0$ is 1. This gives $|\sqrt{3}h + k - 6| = 2$. Since the origin and center are on the same side of the line, $\sqrt{3}h + k - 6 = -2$, so $\sqrt{3}h + k = 4$. The point D is $(\frac{3\sqrt{3}}{2},\frac{3}{2})$. The radius OD is perpendicular to PQ. The slope of PQ is $-\sqrt{3}$. The slope of OD is $\frac{k - 3/2}{h - 3\sqrt{3}/2} = \frac{1}{\sqrt{3}}$. This simplifies to $\sqrt{3}k - \frac{3\sqrt{3}}{2} = h - \frac{3\sqrt{3}}{2}$, so $h = \sqrt{3}k$. Substituting into $\sqrt{3}h + k = 4$, we get $\sqrt{3}(\sqrt{3}k) + k = 4$, which gives $4k = 4$, so $k=1$. Then $h=\sqrt{3}$. The center is $(\sqrt{3}, 1)$. The equation is $(x - \sqrt{3})^2 + (y - 1)^2 = 1$.