The area (in sq. units) of the part of circle $x^{2}+y^{2}=169$ which is below the line $5x-y=13$ is $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin^{-1}\left(\frac{12}{13}\right)$ where $\alpha, \beta$ are coprime numbers. Then $\alpha+\beta$ is equal to

Area Under Curves — JEE Main · 2024 · 29 Jan 2024 · Shift 1 Question with Solution

The area (in sq. units) of the part of circle

x^{2} + y^{2} = 169

which is below the line

5 x - y = 13

\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} sin^{- 1} (\frac{12}{13})

where

α, β

are coprime numbers. Then

α + β

is equal to

Answer: 171

The line

5 x - y = 13

meets the circle

x^{2} + y^{2} = 169

(radius 13) at

(5, 12)

and

(0, - 13)

(both satisfy the line and the circle).

The central angle

θ

between these two points:

cos θ = \frac{O A \cdot O B}{169} = \frac{0 - 156}{169} = - \frac{12}{13}

, so

θ = \frac{π}{2} + sin^{- 1} (\frac{12}{13})

(using

cos θ = - sin (θ - π /2)

and

sin θ = \frac{5}{13}

).

The area of the circular segment cut off by this chord (the region below the line, containing the origin's opposite side) is

Area = \frac{1}{2} r^{2} (θ - sin θ) = \frac{169}{2} [\frac{π}{2} + sin^{- 1} (\frac{12}{13}) - \frac{5}{13}]

= \frac{169 π}{4} + \frac{169}{2} sin^{- 1} (\frac{12}{13}) - \frac{65}{2}

Matching this to the given form

\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} sin^{- 1} (\frac{12}{13})

gives

α = 169, β = 2

(coprime, since

169 = 1 3^{2}

α + β = 171

Key confirmed: 171.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer