JEE MainMathematics

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Area Bounded Between a Circle and an Absolute Value Curve

Q: How do we know the angle subtended at the origin is exactly 90 degrees?

The vector from the origin to (-3, 4) is u vector = (-3, 4) and to (4, 3) is v vector = (4, 3). The dot product is u vector·v vector = (-3)(4) + (4)(3) = 0, so the vectors are orthogonal.

If the area of the larger portion bounded between the curves x^2 + y^2 = 25 and y = |x - 1| is 1/4(bπ + c), b, c ∈ ℕ, then b + c is equal to

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Correct answer

The area of the larger region is $\frac{1}{4} (75 π + 2)$ , which gives $b = 75$ , $c = 2$ , and $b + c = 77$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Circle: $x^{2} + y^{2} = 25$ of radius $R = 5$ . V-shaped curve: $y = ∣ x - 1 ∣$ .

02Goal

Find the area $A$ of the larger portion between the two curves and write it in the form $\frac{1}{4} (b π + c)$ to find $b + c$ .

03Approach

Find intersection points of $y = ∣ x - 1 ∣$ with $x^{2} + y^{2} = 25$ . For $x \geq 1$ , $x^{2} + (x - 1)^{2} = 25 ⟹ 2 x^{2} - 2 x - 24 = 0 ⟹ x^{2} - x - 12 = 0 ⟹ (x - 4) (x + 3) = 0 ⟹ x = 4$ , $y = 3$ . For $x < 1$ , $x^{2} + (1 - x)^{2} = 25 ⟹ x = - 3$ , $y = 4$ . Notice the chords from the origin to $(4, 3)$ and $(- 3, 4)$ are mutually perpendicular, as $(4) (- 3) + (3) (4) = 0$ . The angle subtended by the arc at the center is $\frac{π}{2}$ .

04Execute

The region bounded inside the V-curve and circle consists of the circular sector subtended by the arc between $(- 3, 4)$ and $(4, 3)$ plus/minus triangles, or we can find the smaller area $A_{small}$ directly: $A_{small} = Area of sector subtending \frac{π}{2} - Area of Δ O (- 3, 4) (4, 3) + Area of Δ (1, 0) (- 3, 4) (4, 3)$ . Alternatively, compute the area under the V-shape above the x-axis and the circular segments: Area under $y = ∣ x - 1 ∣$ between $x = - 3$ and $x = 4$ is $\int_{- 3}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x = \frac{1}{2} (4) (4) + \frac{1}{2} (3) (3) = 8 + 4.5 = 12.5 = \frac{25}{2}$ . The area of the sector of the circle from $θ_{1} = \arccos (- 3 / 5)$ to $θ_{2} = \arccos (4 / 5)$ has angle $Δ θ = \frac{π}{2}$ . The area of the minor segment bounded by the chord connecting $(- 3, 4)$ and $(4, 3)$ is $\frac{1}{2} R^{2} (\frac{π}{2} - \sin \frac{π}{2}) = \frac{25}{2} (\frac{π}{2} - 1) = \frac{25 π}{4} - \frac{25}{2}$ . The smaller area is $A_{small} = Area of minor segment + Area of triangle with vertices (- 3, 4), (4, 3), (1, 0)$ . Area of triangle with vertices $(1, 0), (4, 3), (- 3, 4)$ : $Area = \frac{1}{2} ∣ 1 (3 - 4) + 4 (4 - 0) + (- 3) (0 - 3) ∣ = \frac{1}{2} ∣ - 1 + 16 + 9 ∣ = \frac{24}{2} = 12$ . Thus, $A_{small} = (\frac{25 π}{4} - \frac{25}{2}) + 12 = \frac{25 π}{4} - \frac{1}{2} = \frac{25 π - 2}{4}$ . The total area of the circle is $π R^{2} = 25 π$ . Therefore, the larger area is $A_{large} = 25 π - A_{small} = 25 π - \frac{25 π - 2}{4} = \frac{75 π + 2}{4}$ . Thus, $b = 75$ and $c = 2$ , which gives $b + c = 75 + 2 = 77$ .

✓Verify

Verify: $A_{large} + A_{small} = \frac{75 π + 2}{4} + \frac{25 π - 2}{4} = \frac{100 π}{4} = 25 π$ , which equals the area of the entire circle. Hence, the partition is fully consistent.

Hints that build this answer step by step

What are the intersection points of the circle $x^{2} + y^{2} = 25$ and the two branches of $y = ∣ x - 1 ∣$ ?
$(- 3, 4)$ on the left branch and $(4, 3)$ on the right branch
What is the angle subtended by the chord connecting $A (4, 3)$ and $B (- 3, 4)$ at the origin $O (0, 0)$ ?
$\frac{π}{2}$ radians, because the dot product of $\vec{O A} = ⟨ 4, 3 ⟩$ and $\vec{O B} = ⟨ - 3, 4 ⟩$ is zero
What is the area of the smaller region bounded between the circle and $y = ∣ x - 1 ∣$ ?
$\frac{25 π - 2}{4}$
Subtracting the smaller area from the total area of the circle ( $25 π$ ), what is the value of $b + c$ ?
$77$

✓ Source and academic review

Question type: Numerical
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

How do we know the angle subtended at the origin is exactly 90 degrees?

The vector from the origin to $(- 3, 4)$ is $\vec{u} = (- 3, 4)$ and to $(4, 3)$ is $\vec{v} = (4, 3)$ . The dot product is $\vec{u} \cdot \vec{v} = (- 3) (4) + (4) (3) = 0$ , so the vectors are orthogonal.

Answer

Accepted answer: 77

The area of the larger region is $\frac{1}{4} (75 π + 2)$ , which gives $b = 75$ , $c = 2$ , and $b + c = 77$ .

Why each option works or fails

Step-by-step solution

Given: Circle: $x^{2} + y^{2} = 25$ of radius $R = 5$ . V-shaped curve: $y = ∣ x - 1 ∣$ .
Goal: Find the area $A$ of the larger portion between the two curves and write it in the form $\frac{1}{4} (b π + c)$ to find $b + c$ .
Approach: Find intersection points of $y = ∣ x - 1 ∣$ with $x^{2} + y^{2} = 25$ . For $x \geq 1$ , $x^{2} + (x - 1)^{2} = 25 ⟹ 2 x^{2} - 2 x - 24 = 0 ⟹ x^{2} - x - 12 = 0 ⟹ (x - 4) (x + 3) = 0 ⟹ x = 4$ , $y = 3$ . For $x < 1$ , $x^{2} + (1 - x)^{2} = 25 ⟹ x = - 3$ , $y = 4$ . Notice the chords from the origin to $(4, 3)$ and $(- 3, 4)$ are mutually perpendicular, as $(4) (- 3) + (3) (4) = 0$ . The angle subtended by the arc at the center is $\frac{π}{2}$ .
Execute: The region bounded inside the V-curve and circle consists of the circular sector subtended by the arc between $x^{2} + (x - 1)^{2} = 25 ⟹ 2 x^{2} - 2 x - 24 = 0 ⟹ x^{2} - x - 12 = 0 ⟹ (x - 4) (x + 3) = 0 ⟹ x = 4$ and $(4, 3)$ plus/minus triangles, or we can find the smaller area $A_{small}$ directly: $A_{small} = Area of sector subtending \frac{π}{2} - Area of Δ O (- 3, 4) (4, 3) + Area of Δ (1, 0) (- 3, 4) (4, 3)$ . Alternatively, compute the area under the V-shape above the x-axis and the circular segments: Area under $y = ∣ x - 1 ∣$ between $x = - 3$ and $x = 4$ is $\int_{- 3}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x = \frac{1}{2} (4) (4) + \frac{1}{2} (3) (3) = 8 + 4.5 = 12.5 = \frac{25}{2}$ . The area of the sector of the circle from $θ_{1} = \arccos (- 3 / 5)$ to $θ_{2} = \arccos (4 / 5)$ has angle $Δ θ = \frac{π}{2}$ . The area of the minor segment bounded by the chord connecting $(- 3, 4)$ and $(4, 3)$ is $\frac{1}{2} R^{2} (\frac{π}{2} - \sin \frac{π}{2}) = \frac{25}{2} (\frac{π}{2} - 1) = \frac{25 π}{4} - \frac{25}{2}$ . The smaller area is $A_{small} = Area of minor segment + Area of triangle with vertices (- 3, 4), (4, 3), (1, 0)$ . Area of triangle with vertices $(1, 0), (4, 3), (- 3, 4)$ : $Area = \frac{1}{2} ∣ 1 (3 - 4) + 4 (4 - 0) + (- 3) (0 - 3) ∣ = \frac{1}{2} ∣ - 1 + 16 + 9 ∣ = \frac{24}{2} = 12$ . Thus, $A_{small} = (\frac{25 π}{4} - \frac{25}{2}) + 12 = \frac{25 π}{4} - \frac{1}{2} = \frac{25 π - 2}{4}$ . The total area of the circle is $π R^{2} = 25 π$ . Therefore, the larger area is $A_{large} = 25 π - A_{small} = 25 π - \frac{25 π - 2}{4} = \frac{75 π + 2}{4}$ . Thus, $b = 75$ and $c = 2$ , which gives $b + c = 75 + 2 = 77$ .
Verify: Verify: $A_{large} + A_{small} = \frac{75 π + 2}{4} + \frac{25 π - 2}{4} = \frac{100 π}{4} = 25 π$ , which equals the area of the entire circle. Hence, the partition is fully consistent.

Area Bounded Between a Circle and an Absolute Value Curve

What feels right?

The area of the larger region is $\frac{1}{4} (75 π + 2)$ , which gives $b = 75$ , $c = 2$ , and $b + c = 77$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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