JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Ratio of Areas Bounded by a Line and a Semicircle

Q: Why is the area under the circle from x = 1 to x = 3 a quarter of a circle without needing explicit integration?

The circle is centered at (1, 0) and has radius 2. Thus, x = 1 is the vertical centerline of the circle. Also, x = 3 is its rightmost edge. In the upper half-plane, the region between x = 1 and x = 3 is the first quadrant of this circle. This region is one quarter of the full circle.

Let A = {(x, y) ∈ ℝ^2: y ≥ 0, 2x ≤ y ≤ √(4 - (x - 1)^2)} and B = {(x, y) ∈ ℝ × ℝ: 0 ≤ y ≤ min{2x, √(4 - (x - 1)^2)}}. Then the ratio of the area of A to the area of B is

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A(π + 1)/(π - 1)Bπ/(π - 1)C(π - 1)/(π + 1)Dπ/(π + 1)

Step-by-step solutionView

Correct answer

The ratio of the area of region $A$ to the area of region $B$ is $\frac{π - 1}{π + 1}$ .

Option analysis

Why each option works or fails

A · (π + 1)/(π - 1)

Inverting the requested ratio and computing $Area (B) / Area (A)$ instead of $Area (A) / Area (B)$ . Ensure you place the area of region $A$ in the numerator and region $B$ in the denominator as requested by the stem.

B · π/(π - 1)

Omitting the triangular area component when calculating the area of region $B$ , taking it to be simply the remaining circular sector. Account for both the sector of the circle and the area under the line $y = 2 x$ up to the intersection point.

C · (π - 1)/(π + 1)

Correct choice. Area of $A$ is $\frac{π - 1}{2}$ and Area of $B$ is $\frac{π + 1}{2}$ , giving the ratio $\frac{π - 1}{π + 1}$ .

D · π/(π + 1)

Replacing the area of region $A$ with the full quarter-circle area rather than subtracting the triangular region. Subtract the triangle formed under $y = 2 x$ from the sector to find the area of region $A$ precisely.

Reviewed route

Solution

StepWorking

01Given

Region $A = {(x, y) \in R^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - (x - 1)^{2}}}$ and region $B = {(x, y) \in R^{2} : 0 \leq y \leq \min {2 x, \sqrt{4 - (x - 1)^{2}}}$ . The upper boundary is the upper semicircle $(x - 1)^{2} + y^{2} = 4$ with radius $r = 2$ centered at $(1, 0)$ , spanning $x \in [- 1, 3]$ .

02Goal

Find the ratio of the area of region $A$ to the area of region $B$ , i.e., $\frac{Area (A)}{Area (B)}$ .

03Approach

Find the point of intersection between $y = 2 x$ and $(x - 1)^{2} + y^{2} = 4$ in the upper half-plane ( $y \geq 0, x \geq 0$ ). Notice that region $A$ and region $B$ partition the upper semicircle for $x \geq 0$ , plus the part of the semicircle in $x < 0$ belongs to $A$ because $2 x < 0 \leq y$ . Thus, $Area (A) + Area (B) = Area of the upper semicircle = \frac{1}{2} π (2^{2}) = 2 π$ . We can find $Area (B)$ by splitting into a triangle from $x = 0$ to the intersection point and a circular sector from the intersection point to $x = 3$ , then obtain $Area (A) = 2 π - Area (B)$ .

04Execute

Substitute $y = 2 x$ into $(x - 1)^{2} + y^{2} = 4$ : $(x - 1)^{2} + 4 x^{2} = 4 ⟹ 5 x^{2} - 2 x - 3 = 0 ⟹ (5 x + 3) (x - 1) = 0$ Since $y \geq 0$ and $y = 2 x$ , we have $x \geq 0$ , so $x = 1$ . At $x = 1$ , $y = 2 (1) = 2$ .

05Execute

Calculate the area of region $B$ . Region $B$ lies under $y = 2 x$ for $x \in [0, 1]$ and under $y = \sqrt{4 - (x - 1)^{2}}$ for $x \in [1, 3]$ . Area under line: ${Area}_{1} = \frac{1}{2} \times 1 \times 2 = 1$ . Area under circle from $x = 1$ to $x = 3$ : Since $x = 1$ is the center of the circle $(x - 1)^{2} + y^{2} = 4$ , the region from $x = 1$ to $x = 3$ under the upper semicircle is exactly one quarter of the circle. ${Area}_{2} = \frac{1}{4} π (2^{2}) = π$ Therefore, $Area (B) = {Area}_{1} + {Area}_{2} = 1 + π$ .

06Execute

Since $A \cup B$ forms the entire upper semicircle of radius 2 (whose area is $\frac{1}{2} π (2^{2}) = 2 π$ ) and their interiors are disjoint: $Area (A) = 2 π - Area (B) = 2 π - (π + 1) = π - 1$ Thus, the ratio is: $\frac{Area (A)}{Area (B)} = \frac{π - 1}{π + 1}$

✓Verify

Verify that $A$ and $B$ partition the upper semicircle: for any $(x, y)$ with $y \geq 0$ and $(x - 1)^{2} + y^{2} \leq 4$ , if $y \leq 2 x$ , then $y \leq \min (2 x, \sqrt{4 - (x - 1)^{2}})$ , so $(x, y) \in B$ . If $y > 2 x$ , then $2 x < y \leq \sqrt{4 - (x - 1)^{2}}$ , so $(x, y) \in A$ . Their intersection is a set of measure zero (the boundary $y = 2 x$ ). Hence $Area (A) + Area (B) = 2 π$ . Since $π \approx 3.14$ , $Area (B) \approx 4.14 > Area (A) \approx 2.14$ , so the ratio must be $< 1$ , matching $\frac{π - 1}{π + 1} \approx 0.517$ .

Hints that build this answer step by step

What curve forms the upper boundary $y = \sqrt{4 - (x - 1)^{2}}$ ?
The upper half of the circle $(x - 1)^{2} + y^{2} = 4$ centered at $(1, 0)$ with radius $2$ .
At what point in the upper half-plane does the line $y = 2 x$ intersect the circle $(x - 1)^{2} + y^{2} = 4$ ?
$(1, 2)$
What are the individual areas of regions $A$ and $B$ ?
$Area (A) = \frac{π - 1}{2}$ and $Area (B) = \frac{π + 1}{2}$
What is the ratio $\frac{Area (A)}{Area (B)}$ ?
$\frac{π - 1}{π + 1}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the area under the circle from x = 1 to x = 3 a quarter of a circle without needing explicit integration?

The circle is centered at (1, 0) and has radius 2. Thus, x = 1 is the vertical centerline of the circle. Also, x = 3 is its rightmost edge. In the upper half-plane, the region between x = 1 and x = 3 is the first quadrant of this circle. This region is one quarter of the full circle.

Answer

The ratio of the area of region $A$ to the area of region $B$ is $\frac{π - 1}{π + 1}$ .

Why each option works or fails

A: (π + 1)/(π - 1) - Inverting the requested ratio and computing $Area (B) / Area (A)$ instead of $Area (A) / Area (B)$ . Ensure you place the area of region $A$ in the numerator and region $B$ in the denominator as requested by the stem.
B: π/(π - 1) - Omitting the triangular area component when calculating the area of region $B$ , taking it to be simply the remaining circular sector. Account for both the sector of the circle and the area under the line $y = 2 x$ up to the intersection point.
C · correct: (π - 1)/(π + 1) - Correct choice. Area of $A$ is $\frac{π - 1}{2}$ and Area of $B$ is $\frac{π + 1}{2}$ , giving the ratio $\frac{π - 1}{π + 1}$ .
D: π/(π + 1) - Replacing the area of region $A$ with the full quarter-circle area rather than subtracting the triangular region. Subtract the triangle formed under $y = 2 x$ from the sector to find the area of region $A$ precisely.

Step-by-step solution

Given: Region $A = {(x, y) \in R^{2} : y \geq 0, 2 x \leq y \leq \sqrt{4 - (x - 1)^{2}}}$ and region $B = {(x, y) \in R^{2} : 0 \leq y \leq \min {2 x, \sqrt{4 - (x - 1)^{2}}}$ . The upper boundary is the upper semicircle $(x - 1)^{2} + y^{2} = 4$ with radius $r = 2$ centered at $(1, 0)$ , spanning $x \in [- 1, 3]$ .
Goal: Find the ratio of the area of region $A$ to the area of region $B$ , i.e., $\frac{Area (A)}{Area (B)}$ .
Approach: Find the point of intersection between $y = 2 x$ and $(x - 1)^{2} + y^{2} = 4$ in the upper half-plane ( $y \geq 0, x \geq 0$ ). Notice that region $A$ and region $B$ partition the upper semicircle for $x \geq 0$ , plus the part of the semicircle in $x < 0$ belongs to $A$ because $2 x < 0 \leq y$ . Thus, $Area (A) + Area (B) = Area of the upper semicircle = \frac{1}{2} π (2^{2}) = 2 π$ . We can find $Area (B)$ by splitting into a triangle from $x = 0$ to the intersection point and a circular sector from the intersection point to $x = 3$ , then obtain $A$ .
Execute: Substitute $y = 2 x$ into $(x - 1)^{2} + y^{2} = 4$ : $(x - 1)^{2} + 4 x^{2} = 4 ⟹ 5 x^{2} - 2 x - 3 = 0 ⟹ (5 x + 3) (x - 1) = 0$ Since $y \geq 0$ and $y = 2 x$ , we have $x \geq 0$ , so $x = 1$ . At $x = 1$ , $y = 2 (1) = 2$ .
Execute: Calculate the area of region $B$ . Region $B$ lies under $y = 2 x$ for $x \in [0, 1]$ and under $y = \sqrt{4 - (x - 1)^{2}}$ for $x \in [1, 3]$ . Area under line: ${Area}_{1} = \frac{1}{2} \times 1 \times 2 = 1$ . Area under circle from $x = 1$ to $x = 3$ : Since $x = 1$ is the center of the circle $(x - 1)^{2} + y^{2} = 4$ , the region from $x \geq 0$ to $x = 1$ under the upper semicircle is exactly one quarter of the circle. $x = 1$ Therefore, $Area (B) = {Area}_{1} + {Area}_{2} = 1 + π$ .
Execute: Since $A \cup B$ forms the entire upper semicircle of radius 2 (whose area is $\frac{1}{2} π (2^{2}) = 2 π$ ) and their interiors are disjoint: $Area (A) = 2 π - Area (B) = 2 π - (π + 1) = π - 1$ Thus, the ratio is: $\frac{Area (A)}{Area (B)} = \frac{π - 1}{π + 1}$
Verify: Verify that $A$ and $B$ partition the upper semicircle: for any $(x, y)$ with $y \geq 0$ and $(x - 1)^{2} + y^{2} \leq 4$ , if $y \leq 2 x$ , then $y \leq \min (2 x, \sqrt{4 - (x - 1)^{2}})$ , so $(x, y) \in B$ . If $y > 2 x$ , then $2 x < y \leq \sqrt{4 - (x - 1)^{2}}$ , so $(x, y) \in A$ . Their intersection is a set of measure zero (the boundary $y = 2 x$ ). Hence $Area (A) + Area (B) = 2 π$ . Since $π \approx 3.14$ , $Area (B) \approx 4.14 > Area (A) \approx 2.14$ , so the ratio must be $< 1$ , matching $\frac{π - 1}{π + 1} \approx 0.517$ .

Ratio of Areas Bounded by a Line and a Semicircle

What feels right?

The ratio of the area of region $A$ to the area of region $B$ is $\frac{π - 1}{π + 1}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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