1. **Problem Statement:** Find the largest area of the region bounded by the circle given by the equation $$x^2 + y^2 = a^2$$ and the vertical line $$y = c$$ where $$-a < c < 0$$. 2. **Understanding the problem:** The circle is centered at the origin with radius $$a$$. The line $$y = c$$ is horizontal (not vertical as mentioned) and lies below the x-axis since $$c$$ is negative. We want the area of the region inside the circle and above the line $$y = c$$. 3. **Formula and approach:** The area bounded by the circle and the line $$y = c$$ is the area of the circular segment above $$y = c$$. The circle equation can be rewritten as $$x = \pm \sqrt{a^2 - y^2}$$. The area of the segment from $$y = c$$ to $$y = a$$ is: $$\text{Area} = \int_c^a 2\sqrt{a^2 - y^2} \, dy$$ 4. **Calculate the integral:** Use the formula for the integral: $$\int \sqrt{a^2 - y^2} \, dy = \frac{y}{2} \sqrt{a^2 - y^2} + \frac{a^2}{2} \arcsin\left(\frac{y}{a}\right) + C$$ So, $$\text{Area} = 2 \left[ \frac{y}{2} \sqrt{a^2 - y^2} + \frac{a^2}{2} \arcsin\left(\frac{y}{a}\right) \right]_c^a$$ 5. **Evaluate at bounds:** At $$y = a$$: $$\frac{a}{2} \sqrt{a^2 - a^2} + \frac{a^2}{2} \arcsin(1) = 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{a^2 \pi}{4}$$ At $$y = c$$: $$\frac{c}{2} \sqrt{a^2 - c^2} + \frac{a^2}{2} \arcsin\left(\frac{c}{a}\right)$$ 6. **Final area expression:** $$\text{Area} = 2 \left( \frac{a^2 \pi}{4} - \left[ \frac{c}{2} \sqrt{a^2 - c^2} + \frac{a^2}{2} \arcsin\left(\frac{c}{a}\right) \right] \right)$$ Simplify: $$\text{Area} = \frac{a^2 \pi}{2} - c \sqrt{a^2 - c^2} - a^2 \arcsin\left(\frac{c}{a}\right)$$ 7. **Maximizing the area:** Since $$c$$ is fixed in $$(-a,0)$$, the area depends on $$c$$. To find the largest area, consider the derivative with respect to $$c$$ and set it to zero. Calculate $$\frac{d}{dc} \text{Area}$$: $$\frac{d}{dc} \text{Area} = - \sqrt{a^2 - c^2} - c \cdot \frac{-c}{\sqrt{a^2 - c^2}} - a^2 \cdot \frac{1}{\sqrt{a^2 - c^2}} \cdot \frac{1}{a}$$ Simplify derivative: $$= - \sqrt{a^2 - c^2} + \frac{c^2}{\sqrt{a^2 - c^2}} - \frac{a}{\sqrt{a^2 - c^2}}$$ Combine terms: $$= \frac{- (a^2 - c^2) + c^2 - a^2}{\sqrt{a^2 - c^2}} = \frac{-a^2 + c^2 + c^2 - a^2}{\sqrt{a^2 - c^2}} = \frac{2c^2 - 2a^2}{\sqrt{a^2 - c^2}}$$ Set derivative to zero: $$2c^2 - 2a^2 = 0 \Rightarrow c^2 = a^2 \Rightarrow c = \pm a$$ But $$c$$ is in $$(-a,0)$$, so $$c = -a$$. 8. **Check endpoints:** At $$c = -a$$, the area is: $$\text{Area} = \frac{a^2 \pi}{2} - (-a) \sqrt{a^2 - a^2} - a^2 \arcsin(-1) = \frac{a^2 \pi}{2} - 0 + a^2 \cdot \frac{-\pi}{2} = \frac{a^2 \pi}{2} + \frac{a^2 \pi}{2} = a^2 \pi$$ This is the area of the entire circle. At $$c = 0$$, the area is: $$\text{Area} = \frac{a^2 \pi}{2} - 0 - a^2 \arcsin(0) = \frac{a^2 \pi}{2}$$ 9. **Conclusion:** The largest area bounded by the circle and the line $$y = c$$ for $$-a < c < 0$$ is when $$c \to -a$$, which is the entire circle area $$a^2 \pi$$. **Final answer:** $$\boxed{\text{Largest area} = a^2 \pi}$$