To find the area common to the circles given by the equations:

1. $x^2 + y^2 = a^2$ (center at origin and radius $a$),
2. $x^2 + y^2 = 2ax$ (center at $(a, 0)$ and radius $a$),

we proceed as follows:

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Step 1: Points of Intersection

Rewriting the second circle’s equation: $x^2 + y^2 = 2ax \implies x^2 - 2ax + y^2 = 0 \implies (x - a)^2 + y^2 = a^2.$ Thus, this circle is centered at $(a, 0)$ with radius $a$. The two circles intersect when: $x^2 + y^2 = a^2 \quad \text{and} \quad (x-a)^2 + y^2 = a^2.$ Expanding $(x-a)^2 + y^2 = a^2$: $x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 - 2ax + y^2 = 0.$ Subtracting $x^2 + y^2 = a^2$: $-2ax + a^2 - a^2 = 0 \implies x = \frac{a}{2}.$ Substitute $x = \frac{a}{2}$ into $x^2 + y^2 = a^2$: $\left(\frac{a}{2}\right)^2 + y^2 = a^2 \implies \frac{a^2}{4} + y^2 = a^2 \implies y^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}.$ $y = \pm \frac{\sqrt{3}a}{2}.$ Thus, the points of intersection are: $\left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right) \quad \text{and} \quad \left(\frac{a}{2}, -\frac{\sqrt{3}a}{2}\right).$

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Step 2: Symmetry and Geometry

The two circles intersect symmetrically, and the common area consists of two equal lens-shaped regions. Using symmetry, we calculate the area of one such region and double it.

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Step 3: Area of Intersection (Lens Formula)

The area of intersection can be computed as: $\text{Common Area} = 2 \cdot \left[\text{Sector Area in Circle 1} - \text{Triangle Area}\right].$

1. Sector Area in Circle 1

The angle subtended at the center of Circle 1 (at the origin) by the points of intersection is $\theta$. Using the dot product: $\cos \theta = \frac{a/2}{a} = \frac{1}{2}.$ Thus, $\theta = \frac{\pi}{3}$. The sector area is: $\text{Sector Area} = \frac{\theta}{2\pi} \cdot \pi a^2 = \frac{\pi a^2}{6}.$

2. Triangle Area

The base of the triangle is the chord length, and the height is $y$-coordinate of the intersection points. The chord length is: $\text{Chord Length} = 2 \cdot \left|y\right| = 2 \cdot \frac{\sqrt{3}a}{2} = \sqrt{3}a.$ The triangle area is: $\text{Triangle Area} = \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} \cdot \sqrt{3}a \cdot \frac{a}{2} = \frac{\sqrt{3}a^2}{4}.$

3. Area of One Lens

The area of one lens (half of the intersection) is: $\text{Lens Area} = \text{Sector Area} - \text{Triangle Area} = \frac{\pi a^2}{6} - \frac{\sqrt{3}a^2}{4}.$

4. Total Common Area

The total common area is: $\text{Common Area} = 2 \cdot \left(\frac{\pi a^2}{6} - \frac{\sqrt{3}a^2}{4}\right) = \frac{\pi a^2}{3} - \frac{\sqrt{3}a^2}{2}.$

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Final Answer:

$\text{Common Area} = \frac{\pi a^2}{3} - \frac{\sqrt{3}a^2}{2}.$

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Would you like further details or a graphical representation?

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Related Questions:

1. How do you derive the equation of a chord given two circles' equations?
2. Can you find the common area for two circles with different radii?
3. What is the formula for the area of a segment of a circle?
4. How do you generalize this method for three intersecting circles?
5. What is the geometric significance of the intersection points?

Tip:

Visualizing problems involving circle intersections helps significantly. Try sketching the circles and their intersection region to better understand the symmetry and relationships.