1. The problem is to find the values of $X$ and $y$ from the equation $$X^2 + y^2 = (1 + xi)(3 + i)$$ where $i$ is the imaginary unit. 2. First, expand the right-hand side by multiplying the complex numbers: $$ (1 + xi)(3 + i) = 1\times 3 + 1\times i + xi \times 3 + xi \times i = 3 + i + 3xi + x i^2 $$ 3. Recall that $i^2 = -1$, so substitute it: $$3 + i + 3xi - x$$ 4. Now group real and imaginary parts: Real part: $$3 - x$$ Imaginary part: $$i + 3xi = i(1 + 3x)$$ 5. The left side of the original equation $X^2 + y^2$ must equal the right side complex expression. Since $X^2 + y^2$ is a real number (sum of squares), the imaginary part must be zero: $$1 + 3x = 0$$ 6. Solve for $x$: $$x = -\frac{1}{3}$$ 7. Substitute $x$ into the real part: $$3 - x = 3 - \left(-\frac{1}{3}\right) = 3 + \frac{1}{3} = \frac{10}{3}$$ 8. Therefore, $$X^2 + y^2 = \frac{10}{3}$$ 9. The problem reduces to the equation of a circle with radius squared $\frac{10}{3}$. Final answers: - $x = -\frac{1}{3}$ - $X^2 + y^2 = \frac{10}{3}$ (for all points on the circle)