1. **Problem Statement:** Find the equation of the circle that passes through the points of intersection of the two given circles $$S_1: x^2 + y^2 + 2x + 3y - 7 = 0$$ and $$S_2: x^2 + y^2 + 8x - 2y - 1 = 0$$ and also passes through the point $(1,2)$. 2. **Key Idea:** The family of circles passing through the intersection points of $S_1$ and $S_2$ can be written as $$S = S_1 + \lambda S_2 = 0$$ where $\lambda$ is a parameter. 3. **Write the combined equation:** $$x^2 + y^2 + 2x + 3y - 7 + \lambda(x^2 + y^2 + 8x - 2y - 1) = 0$$ Simplify: $$ (1 + \lambda)(x^2 + y^2) + (2 + 8\lambda)x + (3 - 2\lambda)y - (7 + \lambda) = 0$$ 4. **Since this is a circle, the coefficient of $x^2$ and $y^2$ must be equal and nonzero.** Here, they are both $1 + \lambda$, so for a circle, $1 + \lambda \neq 0$. 5. **Use the point $(1,2)$ to find $\lambda$:** Substitute $x=1$, $y=2$: $$ (1 + \lambda)(1^2 + 2^2) + (2 + 8\lambda)(1) + (3 - 2\lambda)(2) - (7 + \lambda) = 0$$ Calculate step-by-step: $$ (1 + \lambda)(1 + 4) + (2 + 8\lambda) + 2(3 - 2\lambda) - 7 - \lambda = 0$$ $$ 5(1 + \lambda) + 2 + 8\lambda + 6 - 4\lambda - 7 - \lambda = 0$$ $$ 5 + 5\lambda + 2 + 8\lambda + 6 - 4\lambda - 7 - \lambda = 0$$ Combine like terms: $$ (5 + 2 + 6 - 7) + (5\lambda + 8\lambda - 4\lambda - \lambda) = 0$$ $$ 6 + 8\lambda = 0$$ 6. **Solve for $\lambda$:** $$ 8\lambda = -6 \implies \lambda = -\frac{6}{8} = -\frac{3}{4}$$ 7. **Write the equation of the required circle:** Substitute $\lambda = -\frac{3}{4}$ back into the combined equation: $$ (1 - \frac{3}{4})(x^2 + y^2) + (2 + 8(-\frac{3}{4}))x + (3 - 2(-\frac{3}{4}))y - (7 - \frac{3}{4}) = 0$$ Simplify coefficients: $$ \frac{1}{4}(x^2 + y^2) + (2 - 6)x + (3 + \frac{3}{2})y - \frac{25}{4} = 0$$ $$ \frac{1}{4}(x^2 + y^2) - 4x + \frac{9}{2}y - \frac{25}{4} = 0$$ Multiply through by 4 to clear denominators: $$ x^2 + y^2 - 16x + 18y - 25 = 0$$ **Final answer:** $$\boxed{x^2 + y^2 - 16x + 18y - 25 = 0}$$