1. **State the problem:** Find the area of the common region bounded by the circle $$(x-2)^2+(y-2)^2=4$$, the line $$x+y=2$$, and the ellipse $$\frac{x^2}{9}+\frac{y^2}{4}=1$$. 2. **Analyze each curve:** - Circle: center at $(2,2)$ with radius $2$. - Line: rearranged as $$y=2-x$$. - Ellipse: centered at origin with semi-major axis $3$ (along x-axis) and semi-minor axis $2$ (along y-axis). 3. **Find intersection points:** - Between circle and line: Substitute $$y=2-x$$ into circle: $$ (x-2)^2 + (2 - x - 2)^2 = 4 $$ $$ (x-2)^2 + (-x)^2 = 4 $$ $$ (x-2)^2 + x^2 = 4 $$ $$ x^2 - 4x + 4 + x^2 = 4 $$ $$ 2x^2 - 4x + 4 = 4 $$ $$ 2x^2 - 4x = 0 $$ $$ 2x(x - 2) = 0 $$ So, $$x=0$$ or $$x=2$$. For $$x=0$$, $$y=2-0=2$$. For $$x=2$$, $$y=2-2=0$$. Intersection points: $(0,2)$ and $(2,0)$. - Between circle and ellipse: Substitute $$y$$ from circle into ellipse or vice versa is complicated; instead, check if points $(0,2)$ and $(2,0)$ lie on ellipse: For $(0,2)$: $$\frac{0^2}{9} + \frac{2^2}{4} = 0 + 1 = 1$$, lies on ellipse. For $(2,0)$: $$\frac{2^2}{9} + \frac{0^2}{4} = \frac{4}{9} + 0 = \frac{4}{9} < 1$$, inside ellipse. - Between line and ellipse: Substitute $$y=2-x$$ into ellipse: $$\frac{x^2}{9} + \frac{(2-x)^2}{4} = 1$$ Multiply both sides by 36 (LCM of 9 and 4): $$4x^2 + 9(2 - x)^2 = 36$$ $$4x^2 + 9(4 - 4x + x^2) = 36$$ $$4x^2 + 36 - 36x + 9x^2 = 36$$ $$13x^2 - 36x + 36 = 36$$ $$13x^2 - 36x = 0$$ $$x(13x - 36) = 0$$ $$x=0$$ or $$x=\frac{36}{13} \approx 2.77$$ For $$x=0$$, $$y=2$$ (point $(0,2)$ already found). For $$x=2.77$$, $$y=2-2.77 = -0.77$$. 4. **Determine the common region:** - The common region is bounded by the circle, line, and ellipse, intersecting at points $(0,2)$ and $(2,0)$. 5. **Set up integral for area:** - The area is the integral over $$x$$ from 0 to 2 of the minimum of the upper boundaries minus the maximum of the lower boundaries. - Upper boundary is the circle or ellipse, lower boundary is the line. 6. **Express $$y$$ from circle:** $$ (x-2)^2 + (y-2)^2 = 4 $$ $$ (y-2)^2 = 4 - (x-2)^2 $$ $$ y-2 = \pm \sqrt{4 - (x-2)^2} $$ Upper semicircle: $$ y = 2 + \sqrt{4 - (x-2)^2} $$ Lower semicircle: $$ y = 2 - \sqrt{4 - (x-2)^2} $$ 7. **Express $$y$$ from ellipse:** $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ $$ y^2 = 4(1 - \frac{x^2}{9}) = 4 - \frac{4x^2}{9} $$ $$ y = \pm \sqrt{4 - \frac{4x^2}{9}} = \pm \frac{2}{3} \sqrt{9 - x^2} $$ 8. **Determine which curve is upper between 0 and 2:** - At $$x=1$$: Circle upper: $$2 + \sqrt{4 - (1-2)^2} = 2 + \sqrt{4 - 1} = 2 + \sqrt{3} \approx 3.732$$ Ellipse upper: $$\frac{2}{3} \sqrt{9 - 1} = \frac{2}{3} \sqrt{8} \approx 1.885$$ So circle upper is above ellipse upper. 9. **Line $$y=2-x$$ is below both curves between 0 and 2.** 10. **Area of common region:** - The region bounded by line, circle, and ellipse between $$x=0$$ and $$x=2$$ is the area under the circle upper curve minus the area under the line, but also bounded by the ellipse. - Since ellipse is inside circle in this interval, the common region is bounded by line, ellipse, and circle. 11. **Calculate area between line and ellipse from $$x=0$$ to $$x=2$$:** $$ A_1 = \int_0^2 \left( \frac{2}{3} \sqrt{9 - x^2} - (2 - x) \right) dx $$ 12. **Calculate area between circle and ellipse from $$x=0$$ to $$x=2$$:** $$ A_2 = \int_0^2 \left( 2 + \sqrt{4 - (x-2)^2} - \frac{2}{3} \sqrt{9 - x^2} \right) dx $$ 13. **Total common area:** $$ A = A_1 + A_2 = \int_0^2 \left( 2 + \sqrt{4 - (x-2)^2} - (2 - x) \right) dx $$ 14. **Simplify integrand:** $$ 2 + \sqrt{4 - (x-2)^2} - 2 + x = x + \sqrt{4 - (x-2)^2} $$ 15. **Final integral:** $$ A = \int_0^2 \left( x + \sqrt{4 - (x-2)^2} \right) dx $$ 16. **Evaluate integral:** - Split integral: $$ \int_0^2 x dx + \int_0^2 \sqrt{4 - (x-2)^2} dx $$ - First integral: $$ \int_0^2 x dx = \left[ \frac{x^2}{2} \right]_0^2 = 2 $$ - Second integral is area under semicircle of radius 2 centered at 2: Substitute $$u = x - 2$$, when $$x=0, u=-2$$; when $$x=2, u=0$$ $$ \int_0^2 \sqrt{4 - (x-2)^2} dx = \int_{-2}^0 \sqrt{4 - u^2} du $$ - Area of semicircle from $$-2$$ to $$0$$ is quarter circle area: $$ \int_{-2}^0 \sqrt{4 - u^2} du = \frac{\pi \times 2^2}{4} = \pi $$ 17. **Sum results:** $$ A = 2 + \pi $$ **Answer:** The area of the common region is $$2 + \pi$$ square units.