1. **Problem 1:** Find the area of the region enclosed between two circles given by $$x^2 + y^2 = 1$$ and $$(x - 1)^2 + y^2 = 1$$ 2. **Problem 2:** Find the area of the region lying above the x-axis and bounded between the circle $$x^2 + y^2 = 2ax$$ and the parabola $$y^2 = 4ac$$ --- ### Problem 1 Steps: 1. The two circles have radius 1. The first is centered at $(0,0)$ and the second at $(1,0)$. 2. The circles intersect where their equations are equal: $$x^2 + y^2 = (x-1)^2 + y^2$$ Simplify: $$x^2 = (x-1)^2 = x^2 - 2x + 1$$ So, $$x^2 = x^2 - 2x + 1 \implies 2x = 1 \implies x = \frac{1}{2}$$ 3. Substitute $x=\frac{1}{2}$ into the first circle to find $y$: $$\left(\frac{1}{2}\right)^2 + y^2 = 1 \implies y^2 = 1 - \frac{1}{4} = \frac{3}{4}$$ So, $$y = \pm \frac{\sqrt{3}}{2}$$ 4. The area of intersection of two circles of radius $r$ with center distance $d$ is given by: $$A = 2r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}$$ Here, $r=1$, $d=1$. 5. Calculate: $$A = 2 \times 1^2 \cos^{-1}\left(\frac{1}{2}\right) - \frac{1}{2} \sqrt{4 - 1} = 2 \times \frac{\pi}{3} - \frac{1}{2} \times \sqrt{3} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$$ 6. The area enclosed between the two circles is twice the lens area (since the problem states enclosed region), so the total area is: $$2 \times A = 2 \times \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{4\pi}{3} - \sqrt{3}$$ --- ### Problem 2 Steps: 1. The circle equation is: $$x^2 + y^2 = 2ax$$ Rewrite as: $$x^2 - 2ax + y^2 = 0 \implies (x - a)^2 + y^2 = a^2$$ So, circle center is $(a,0)$ and radius $a$. 2. The parabola is: $$y^2 = 4ac$$ This is a horizontal line at $y = \pm 2\sqrt{ac}$. 3. We want the area above the x-axis bounded between the circle and parabola. 4. The parabola is a horizontal line, so the region is between $y=0$ and $y=2\sqrt{ac}$. 5. Solve for $x$ from the circle: $$y^2 = 2ax - x^2 \implies x^2 - 2ax + y^2 = 0$$ For fixed $y$, solve quadratic in $x$: $$x = a \pm \sqrt{a^2 - y^2}$$ 6. The region above x-axis is between $y=0$ and $y=2\sqrt{ac}$. 7. The parabola $y^2=4ac$ is constant in $y$, so $c$ must be such that $2\sqrt{ac} \leq a$ for intersection. 8. The area is integral over $y$ from 0 to $2\sqrt{ac}$ of the horizontal distance between circle and parabola: $$A = \int_0^{2\sqrt{ac}} \left(a + \sqrt{a^2 - y^2} - (a - \sqrt{a^2 - y^2})\right) dy = \int_0^{2\sqrt{ac}} 2\sqrt{a^2 - y^2} dy$$ 9. Evaluate integral: $$\int 2\sqrt{a^2 - y^2} dy = y \sqrt{a^2 - y^2} + a^2 \sin^{-1}\left(\frac{y}{a}\right) + C$$ 10. Substitute limits: $$A = \left[ y \sqrt{a^2 - y^2} + a^2 \sin^{-1}\left(\frac{y}{a}\right) \right]_0^{2\sqrt{ac}}$$ 11. Since $2\sqrt{ac} \leq a$, this is valid. 12. Final area: $$A = 2\sqrt{ac} \sqrt{a^2 - 4ac} + a^2 \sin^{-1}\left(\frac{2\sqrt{ac}}{a}\right)$$ --- **Final answers:** 1. Area between two circles: $$\boxed{\frac{4\pi}{3} - \sqrt{3}}$$ 2. Area above x-axis between circle and parabola: $$\boxed{2\sqrt{ac} \sqrt{a^2 - 4ac} + a^2 \sin^{-1}\left(\frac{2\sqrt{ac}}{a}\right)}$$