Understanding the Problem: Circle and Line Partition

The problem asks us to find the area of the major segment created when a specific circle is divided by a line. The circle is given by the equation \((x - a)^2 + y^2 = a^2\), and the line is \(y = x\).

First, let's understand the properties of the circle:

• The standard equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
• Comparing this to \((x - a)^2 + y^2 = a^2\), we see the center of the circle is \((a, 0)\) and the radius is \(a\).

The line \(y = x\) passes through the origin \((0, 0)\) and has a slope of 1.

Finding Intersection Points of the Circle and Line

To find the points where the line \(y = x\) intersects the circle \((x - a)^2 + y^2 = a^2\), we substitute \(y = x\) into the circle equation:

\[ (x - a)^2 + x^2 = a^2 \]\[ x^2 - 2ax + a^2 + x^2 = a^2 \]\[ 2x^2 - 2ax = 0 \]\[ 2x(x - a) = 0 \]From this equation, we get two possible values for \(x\): \(x = 0\) or \(x = a\).

Since \(y = x\), the corresponding \(y\) values are:

• If \(x = 0\), then \(y = 0\). The intersection point is \((0, 0)\).
• If \(x = a\), then \(y = a\). The intersection point is \((a, a)\).

So, the line \(y = x\) intersects the circle at points P\((0, 0)\) and Q\((a, a)\).

Calculating the Angle Subtended at the Center

The line segment PQ connecting the intersection points is a chord of the circle. This chord divides the circle into two segments: a minor segment and a major segment.

To find the area of these segments, we first need to consider the sector formed by the center of the circle C\((a, 0)\) and the points P\((0, 0)\) and Q\((a, a)\).

Let's find the angle \(\theta\) between the lines CP and CQ at the center C\((a, 0)\).

• Vector \(\vec{CP}\) from C\((a, 0)\) to P\((0, 0)\) is \((0 - a, 0 - 0) = (-a, 0)\).
• Vector \(\vec{CQ}\) from C\((a, 0)\) to Q\((a, a)\) is \((a - a, a - 0) = (0, a)\).

We can find the angle using the dot product:

\[ \vec{CP} \cdot \vec{CQ} = |\vec{CP}| |\vec{CQ}| \cos \theta \]\[ (-a)(0) + (0)(a) = \sqrt{(-a)^2 + 0^2} \sqrt{0^2 + a^2} \cos \theta \]\[ 0 = \sqrt{a^2} \sqrt{a^2} \cos \theta \]\[ 0 = a \cdot a \cos \theta \]\[ 0 = a^2 \cos \theta \]Since \(a\) is the radius and \(a > 0\), we must have \(\cos \theta = 0\). This means the angle \(\theta\) is \(\frac{\pi}{2}\) radians or 90 degrees.

The angle subtended by the chord PQ at the center is \(\frac{\pi}{2}\).

Calculating Area of Sector CPQ

The area of a sector with radius \(r\) and central angle \(\theta\) (in radians) is given by \(\frac{1}{2} r^2 \theta\).

Here, radius \(r = a\) and angle \(\theta = \frac{\pi}{2}\).

Area of Sector CPQ = \(\frac{1}{2} a^2 \left(\frac{\pi}{2}\right) = \frac{\pi a^2}{4}\).

Calculating Area of Triangle CPQ

The triangle CPQ has vertices C\((a, 0)\), P\((0, 0)\), and Q\((a, a)\). We found that the angle at C is 90 degrees, and the lengths of CP and CQ are both \(a\) (the radius).

The area of a right-angled triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Using CP as the base and CQ as the height (since they are perpendicular):

Area of Triangle CPQ = \(\frac{1}{2} \times \text{CP} \times \text{CQ} = \frac{1}{2} \times a \times a = \frac{a^2}{2}\).

Calculating Area of Minor Segment

The area of the minor segment is the area of the sector CPQ minus the area of the triangle CPQ.

Area of Minor Segment = Area of Sector CPQ - Area of Triangle CPQ

Area of Minor Segment = \(\frac{\pi a^2}{4} - \frac{a^2}{2}\)

Area of Minor Segment = \(\frac{\pi a^2 - 2a^2}{4} = \frac{(\pi - 2) a^2}{4}\).

Calculating Area of Major Segment

The total area of the circle is \(\pi r^2 = \pi a^2\).

The area of the major segment is the total area of the circle minus the area of the minor segment.

Area of Major Segment = Total Area of Circle - Area of Minor Segment

Area of Major Segment = \(\pi a^2 - \frac{(\pi - 2) a^2}{4}\)

To subtract, find a common denominator:

Area of Major Segment = \(\frac{4 \pi a^2}{4} - \frac{(\pi - 2) a^2}{4}\)

Area of Major Segment = \(\frac{4 \pi a^2 - (\pi a^2 - 2 a^2)}{4}\)

Area of Major Segment = \(\frac{4 \pi a^2 - \pi a^2 + 2 a^2}{4}\)

Area of Major Segment = \(\frac{(4\pi - \pi) a^2 + 2 a^2}{4}\)

Area of Major Segment = \(\frac{3 \pi a^2 + 2 a^2}{4}\)

Area of Major Segment = \(\frac{(3 \pi + 2) a^2}{4}\).

Final Answer

The area of the major segment is \(\frac{(3 \pi + 2) a^2}{4}\).

Revision Table: Circle Segments and Areas

Let's summarize the key formulas used in calculating the area of circle segments.

Additional Information: Geometry of Circle Segments

A circle segment is a region of a circle which is "cut off" from the rest of the circle by a secant or a chord. The segment does not contain the center of the circle.

• A chord divides the circle into two segments.
• The smaller region is called the minor segment.
• The larger region is called the major segment.
• The area of a segment is found by calculating the area of the sector formed by the radii to the endpoints of the chord and subtracting the area of the triangle formed by the center and the endpoints of the chord.
• In this specific problem, the central angle being \(\frac{\pi}{2}\) (or 90 degrees) simplified the calculation of the triangle area because it was a right-angled triangle.