1. **Problem Statement:** Find the intersection points of the circles given by $$x^2 + y^2 = 1$$ and $$(x - 1)^2 + (y - 2)^2 = 4$$ 2. **Step 1: Write down the equations clearly:** Circle 1: $$x^2 + y^2 = 1$$ Circle 2: $$(x - 1)^2 + (y - 2)^2 = 4$$ 3. **Step 2: Expand the second circle's equation:** $$(x - 1)^2 + (y - 2)^2 = 4$$ Expands to: $$x^2 - 2x + 1 + y^2 - 4y + 4 = 4$$ Simplify: $$x^2 + y^2 - 2x - 4y + 5 = 4$$ $$x^2 + y^2 - 2x - 4y + 1 = 0$$ 4. **Step 3: Subtract the first circle's equation from this:** $$(x^2 + y^2 - 2x - 4y + 1) - (x^2 + y^2) = 0 - 1$$ Simplify: $$-2x - 4y + 1 = -1$$ $$-2x - 4y = -2$$ Divide both sides by -2: $$\cancel{-2}x + \cancel{-4}y = \cancel{-2}$$ $$x + 2y = 1$$ This is the line found by eliminating $x^2$ and $y^2$. 5. **Step 4: Use the line equation $x + 2y = 1$ to find intersection points:** Express $x$ in terms of $y$: $$x = 1 - 2y$$ Substitute into Circle 1 equation: $$x^2 + y^2 = 1$$ $$ (1 - 2y)^2 + y^2 = 1$$ Expand: $$1 - 4y + 4y^2 + y^2 = 1$$ Simplify: $$5y^2 - 4y + 1 = 1$$ Subtract 1 from both sides: $$5y^2 - 4y = 0$$ Factor: $$y(5y - 4) = 0$$ So, $$y = 0$$ or $$5y - 4 = 0 \Rightarrow y = \frac{4}{5}$$ 6. **Step 5: Find corresponding $x$ values:** For $y=0$: $$x = 1 - 2(0) = 1$$ For $y=\frac{4}{5}$: $$x = 1 - 2 \times \frac{4}{5} = 1 - \frac{8}{5} = -\frac{3}{5}$$ 7. **Step 6: Intersection points are:** $$\boxed{(1, 0) \text{ and } \left(-\frac{3}{5}, \frac{4}{5}\right)}$$ 8. **Step 7: Plausibility check by sketch:** - Circle 1 is centered at $(0,0)$ with radius 1. - Circle 2 is centered at $(1,2)$ with radius 2. - The points $(1,0)$ and $(-\frac{3}{5}, \frac{4}{5})$ lie on both circles, consistent with the geometry. 9. **Step 8: Geometric meaning of the line $x + 2y - 1 = 0$:** This line is the radical line of the two circles, representing the locus of points with equal power with respect to both circles. It is the line along which the two circles intersect. --- **Final answers:** Intersection points: $$\boxed{(1, 0) \text{ and } \left(-\frac{3}{5}, \frac{4}{5}\right)}$$ Line of intersection (radical line): $$x + 2y - 1 = 0$$