Finding the Area of the Minor Segment of a Circle

The problem asks for the area of the minor segment created by the line \(y = x\) intersecting the circle \((x - a)^2 + y^2 = a^2\).

Let's first understand the given equations:

• The circle has the equation \((x - a)^2 + y^2 = a^2\). This is a standard form of a circle equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. In this case, the center is \((a, 0)\) and the radius is \(a\).
• The line is \(y = x\). This is a straight line passing through the origin with a slope of 1.

When a line intersects a circle, it generally cuts the circle into two parts called segments. To find the area of a segment, we typically find the area of the sector formed by the radii to the intersection points and the center, and subtract the area of the triangle formed by the radii and the chord connecting the intersection points.

Step 1: Find the Points of Intersection

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

\((x - a)^2 + (x)^2 = a^2\)

\(x^2 - 2ax + a^2 + x^2 = a^2\)

\(2x^2 - 2ax = 0\)

\(2x(x - a) = 0\)

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

Let the intersection points be \(A = (0, 0)\) and \(B = (a, a)\). The center of the circle is \(C = (a, 0)\) and the radius is \(a\).

Step 2: Determine the Angle Subtended at the Center

The minor segment is bounded by the chord AB and the arc AB. To find the area of the sector CAB, we need the angle \(\theta\) between the radii CA and CB at the center C. The points are \(C(a, 0)\), \(A(0, 0)\), and \(B(a, a)\).

Let's consider the vectors from the center \(C\) to the intersection points \(A\) and \(B\):

• Vector \(\vec{CA}\) = \(A - C\) = \((0 - a, 0 - 0) = (-a, 0)\).
• Vector \(\vec{CB}\) = \(B - C\) = \((a - a, a - 0) = (0, a)\).

We can find the angle between these vectors using the dot product formula \(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta\).

• Magnitude of \(\vec{CA}\), \(|\vec{CA}| = \sqrt{(-a)^2 + 0^2} = \sqrt{a^2} = a\) (since radius is \(a > 0\)).
• Magnitude of \(\vec{CB}\), \(|\vec{CB}| = \sqrt{0^2 + a^2} = \sqrt{a^2} = a\).
• Dot product \(\vec{CA} \cdot \vec{CB} = (-a)(0) + (0)(a) = 0\).

Since the dot product is 0, the angle \(\theta\) between \(\vec{CA}\) and \(\vec{CB}\) satisfies \(\cos \theta = \frac{0}{a \cdot a} = 0\). For an angle between 0 and \(\pi\) (which is the range for the angle within a triangle or sector), \(\theta = \frac{\pi}{2}\) radians (or 90 degrees).

Step 3: Calculate the Area of the Sector

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

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

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

Step 4: Calculate the Area of the Triangle

The triangle is formed by the center \(C(a, 0)\) and the intersection points \(A(0, 0)\) and \(B(a, a)\).

We can use the coordinates or recognize the triangle type. The vectors \(\vec{CA}\) and \(\vec{CB}\) are perpendicular and have magnitudes \(a\). This means triangle CAB is a right-angled triangle with legs CA and CB of length \(a\).

The area of a right-angled triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Using CA as the base (length \(a\)) and CB as the height (length \(a\)), the area is:

Area of Triangle CAB = \(\frac{1}{2} \times |\vec{CA}| \times |\vec{CB}| = \frac{1}{2} \times a \times a = \frac{a^2}{2}\).

Alternatively, using the determinant formula for triangle area with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):

Area = \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

Using \(C(a, 0)\), \(A(0, 0)\), \(B(a, a)\):

Area = \(\frac{1}{2} |a(0 - a) + 0(a - 0) + a(0 - 0)| = \frac{1}{2} |a(-a) + 0 + 0| = \frac{1}{2} |-a^2| = \frac{a^2}{2}\) (since \(a^2 \ge 0\)).

Step 5: Calculate the Area of the Minor Segment

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

Area of Segment = Area of Sector CAB - Area of Triangle CAB

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

To combine these terms, find a common denominator:

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

Since the central angle is \(\frac{\pi}{2}\) (which is less than \(\pi\)), the sector forms the minor segment. The calculated area is the area of the minor segment.

The area of the minor segment is \(\frac{(\pi - 2) a^2}{4}\).

Revision Table: Key Concepts in Segment Area Calculation

Additional Information: Major vs. Minor Segment

A chord divides a circle into two segments: a minor segment and a major segment.

• The minor segment is the smaller region. Its area is calculated by taking the sector area corresponding to the smaller central angle and subtracting the triangle area.
• The major segment is the larger region. Its area is calculated by taking the area of the whole circle minus the area of the minor segment. Alternatively, it is the sector area corresponding to the reflex angle (\(2\pi - \theta\)) minus the triangle area, but typically calculated as Total Circle Area - Minor Segment Area.

In this problem, the central angle subtended by the chord at the center is \(\frac{\pi}{2}\) radians. Since \(\frac{\pi}{2} < \pi\), the sector with this angle corresponds to the minor segment. The area we calculated, \(\frac{(\pi - 2) a^2}{4}\), is indeed the area of the minor segment.

The area of the whole circle is \(\pi r^2 = \pi a^2\). The area of the major segment would be \(\pi a^2 - \frac{(\pi - 2) a^2}{4}\).