1. The problem involves two curves intersecting: (ii) a circle centered at $O(0,0)$ with radius 1, and (i) a blue curve intersecting the circle at points $A(\frac{1}{2}, \sqrt{2})$ and $B(\frac{1}{2}, -\sqrt{2})$. 2. First, verify the radius of the circle (ii). The radius $r$ is the distance from the center $O(0,0)$ to any point on the circle, for example $A(\frac{1}{2}, \sqrt{2})$: $$r = \sqrt{\left(\frac{1}{2} - 0\right)^2 + (\sqrt{2} - 0)^2} = \sqrt{\frac{1}{4} + 2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$ 3. This shows the radius is $\frac{3}{2}$, not 1, so the problem statement about radius 1 is inconsistent with the points given. We proceed with radius $\frac{3}{2}$. 4. The equation of the circle (ii) centered at $O(0,0)$ with radius $\frac{3}{2}$ is: $$x^2 + y^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$ 5. The blue curve (i) intersects the circle at points $A$ and $B$ which have the same $x$-coordinate $x=\frac{1}{2}$ and $y=\pm \sqrt{2}$. 6. To find the equation of the blue curve (i), note it passes through $A$ and $B$ and intersects the circle there. Since $A$ and $B$ have the same $x$ but opposite $y$, the blue curve could be a vertical line: $$x = \frac{1}{2}$$ 7. Check if this line intersects the circle at $A$ and $B$: Substitute $x=\frac{1}{2}$ into the circle equation: $$\left(\frac{1}{2}\right)^2 + y^2 = \frac{9}{4} \implies y^2 = \frac{9}{4} - \frac{1}{4} = 2$$ $$y = \pm \sqrt{2}$$ which matches points $A$ and $B$. 8. Therefore, the blue curve (i) is the vertical line $x=\frac{1}{2}$. **Final answers:** - Circle (ii): $$x^2 + y^2 = \frac{9}{4}$$ - Blue curve (i): $$x = \frac{1}{2}$$