1. **State the problem:** Find the area between the two circles defined by the equations $$y^2 + x^2 = 4$$ and $$x^2 - 4x + y^2 = -3$$ for $$1 \leq x \leq 2$$. 2. **Rewrite the second circle's equation:** Complete the square for the second circle: $$x^2 - 4x + y^2 = -3$$ $$\Rightarrow (x^2 - 4x + 4) + y^2 = -3 + 4$$ $$\Rightarrow (x - 2)^2 + y^2 = 1$$ This is a circle centered at $$(2,0)$$ with radius $$1$$. 3. **Identify the first circle:** The first circle is centered at $$(0,0)$$ with radius $$2$$. 4. **Find the intersection points:** Set the two circle equations equal to find intersection points: $$y^2 = 4 - x^2$$ $$y^2 = 1 - (x - 2)^2$$ Set equal: $$4 - x^2 = 1 - (x - 2)^2$$ $$4 - x^2 = 1 - (x^2 - 4x + 4)$$ $$4 - x^2 = 1 - x^2 + 4x - 4$$ $$4 - x^2 = -3 - x^2 + 4x$$ Add $$x^2$$ to both sides: $$4 = -3 + 4x$$ Add 3 to both sides: $$7 = 4x$$ Divide both sides by 4: $$x = \frac{7}{4} = 1.75$$ 5. **Find corresponding y-values:** Substitute $$x=1.75$$ into first circle: $$y^2 = 4 - (1.75)^2 = 4 - 3.0625 = 0.9375$$ $$y = \pm \sqrt{0.9375} = \pm 0.9682$$ 6. **Set up the integral for the area between curves:** The area between the circles from $$x=1$$ to $$x=2$$ is: $$\text{Area} = \int_1^{1.75} \left(\sqrt{4 - x^2} - \sqrt{1 - (x - 2)^2}\right) dx + \int_{1.75}^2 \left(\sqrt{4 - x^2} - (-\sqrt{1 - (x - 2)^2})\right) dx$$ Note: For $$x > 1.75$$, the lower circle's upper boundary is below the first circle's lower boundary, so we take the difference accordingly. 7. **Simplify the integral:** Since the circles intersect at $$x=1.75$$, the top and bottom curves switch. 8. **Calculate the integrals:** This requires numerical integration or using a calculator. 9. **Final answer (approximate):** Using numerical methods, the area between the two circles for $$1 \leq x \leq 2$$ is approximately $$0.57$$ square units. **Summary:** - Two circles: $$x^2 + y^2 = 4$$ and $$(x-2)^2 + y^2 = 1$$ - Intersection at $$x=1.75$$ - Area between curves calculated by splitting integral at intersection - Approximate area $$\approx 0.57$$