Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The area of the smaller part of the circle \(x^{2}+y^{2}=a^{2}\), cut off by the line \(x=\frac{a}{\sqrt{2}}\), is given by (A) \(\frac{a^{2}}{2}\left(\frac{\pi}{2}+1\right)\) (B) \(\frac{a^{2}}{2}\left(\frac{\pi}{2}-1\right)\) (C) \(a^{2}\left(\frac{\pi}{2}-1\right)\) (D) None of these

Step-by-Step Solution

Answer

(B) \( \frac{a^2}{2} \left(\frac{\pi}{2} - 1\right) \).

Step 1: Understand the Circle Equation

The given circle has the equation \( x^2 + y^2 = a^2 \). This represents a circle centered at the origin \((0,0)\) with radius \( a \). The circle is symmetric about both the x-axis and the y-axis.

Step 2: Analyze the Line Equation

The line given is \( x=\frac{a}{\sqrt{2}} \). This is a vertical line that intersects the x-axis at \( \frac{a}{\sqrt{2}} \). This line will cut the circle into two parts, one of which is smaller.

Step 3: Find Intersections of Line and Circle

Substitute \( x=\frac{a}{\sqrt{2}} \) into the circle equation: \( \left(\frac{a}{\sqrt{2}}\right)^2 + y^2 = a^2 \). Simplifying gives \( \frac{a^2}{2} + y^2 = a^2 \). Solving for \( y^2 \) gives \( y^2 = \frac{a^2}{2} \), so \( y = \pm \frac{a}{\sqrt{2}} \). These are the points \( \left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right) \) and \( \left(\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}\right) \).

Step 4: Evaluate Area of Sector

The angle subtended by these points at the center of the circle can be found. These points are symmetrically above and below the x-axis along the vertical line. This creates a 45-degree angle ( \( \frac{\pi}{4} \text{ radians} \)) above and below the x-axis for a total of \( \frac{\pi}{2} \text{ radians} \). The area of this sector is given by \( \frac{1}{2} \times a^2 \times \frac{\pi}{2} \).

Step 5: Calculate Area of Triangle

The line forms a right triangle with the center of the circle at the origin and the x-intercept at \( \left(\frac{a}{\sqrt{2}}, 0\right) \). The legs of this triangle have lengths \( \frac{a}{\sqrt{2}} \) each. Thus, the area of the triangle is \( \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}} = \frac{a^2}{4} \).

Step 6: Determine the Area of the Smaller Part

Subtract the area of the triangle from the area of the sector to get the area of the smaller part: \( \frac{1}{2} a^2 \frac{\pi}{2} - \frac{a^2}{4} = \frac{a^2}{2} \left(\frac{\pi}{2} - 1\right) \).

Step 7: Check the Options

Compare the calculated area of the smaller part of the circle with the options given in the problem. The correct match is option (B): \( \frac{a^2}{2} \left(\frac{\pi}{2} - 1\right) \).

Key Concepts

Circle Equation

In geometry, a circle equation helps us understand the geometric properties of a circle. The standard equation for a circle in a coordinate plane is given by \( x^2 + y^2 = a^2 \). This equation represents a circle with its center at the origin \((0,0)\), and a radius of \( a \). This means that every point \((x,y)\) on the circle is exactly \( a \) units away from the origin.

A key feature of this circle is its symmetry. It is symmetric about both the x-axis and the y-axis. This simply means that if you flip the circle over either axis, it will still look the same. This symmetry simplifies problems involving circles as it reduces the complexity of calculations. Understanding this concept is crucial as it sets the stage for exploring intersections and areas within the circle.

Line Intersection with Circle

The intersection of a line with a circle is a classic problem in coordinate geometry. Given a vertical line like \(x = \frac{a}{\sqrt{2}}\), it intersects the circle \(x^2 + y^2 = a^2\) at points determined by substituting the line equation into the circle equation.

When we plug \( x = \frac{a}{\sqrt{2}} \) into the circle equation, we have:

• Substitution: \( \left(\frac{a}{\sqrt{2}}\right)^2 + y^2 = a^2 \)
• Simplification: \( \frac{a^2}{2} + y^2 = a^2 \)
• Solving for \( y^2 \): \( y^2 = \frac{a^2}{2} \)

This gives us intersection points \( \left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right) \) and \( \left(\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}\right) \). These points are symmetric about the x-axis. Recognizing these intersection points is important as they define the boundary of a segment within the circle that we aim to analyze.

Sector Area

The sector of a circle is a pie-shaped part of the circle, and calculating its area can help in solving various geometrical problems. The sector area is centered around the origin, with lines drawn to the points of intersection with the circle.

To find the sector area, we observe that the line intersects the circle creating two identical right triangles. The angle subtended by these intersections at the circle's center covers half the circle's top half with a total angle of \(\frac{\pi}{2}\) radians. The formula for the area of a sector is given by:

• \( \text{Sector Area} = \frac{1}{2} \, \times \text{radius}^2 \, \times \text{angle in radians} \)
• With the circle having a radius \(a\) and angle \(\frac{\pi}{2}\), the area becomes \( \frac{1}{2} a^2 \frac{\pi}{2} \).

This area calculation is instrumental in determining the area of smaller segments when combined with triangle area computation.

Triangle Area in Coordinate Geometry

Calculating the area of a triangle formed within a circle can provide insights into solving area-related circle problems. In reference to our exercise, a right triangle is formed with its points at the circle's center, the x-intercept, and one of the intersection points.

The legs of this triangle are \( \frac{a}{\sqrt{2}} \) each, forming a 45-degree right triangle. The formula for the area of a right triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \), which simplifies to:

• \( \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}} = \frac{a^2}{4} \)

This triangle area is subtracted from the sector area to find the small segment area of the circle. Understanding this relationship between sector and triangle areas aids in solving geometrical partition problems effectively.