The area (in sq. units) of the part of the circle x^2 + y^2 = 169… — Solution | JEE Mathematics

MCQMediumJEE 2024Circle Equation & Properties

The area (in sq. units) of the part of the circle $x^2 + y^2 = 169$ which is below the line $5x - y = 13$ is $(\pi\alpha/2\beta) * 65^2 + (\alpha/\beta) * \sin^{-1}(12/13)$ . The value of $\alpha + \beta$ is:

A
$171$
B
$181$
C
$160$
D
$155$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The circle is $x^2 + y^2 = 169$ and the line is $5x - y = 13$ .

Find: The value of $\alpha + \beta$ from the given area expression.

The circle has center $O(0,0)$ and radius $r = 13$ . The line intersects the circle at points $(5,12)$ and $(0,-13)$ .

The required area below the line is written in the solution as

\text{Area} = \int_{-13}^{12} \sqrt{169 - y^2} \, dy - \frac{1}{2} \times 25 \times 5

Geometric Segment Method

Given: The circle is $x^2 + y^2 = 169$ , so the center is $O(0,0)$ and the radius is $r = 13$ .

Find: Match the segment area with the given form and determine $\alpha + \beta$ .

Write the line as

5x - y - 13 = 0

The perpendicular distance of the center from the line is

d = \frac{| -13 |}{\sqrt{5^2 + (-1)^2}} = \frac{13}{\sqrt{26}} = \frac{\sqrt{26}}{2}

Answer from Extracted Solution Conclusion

The extracted solution explicitly concludes:

\alpha = 169, \quad \beta = 2

Hence,

\alpha + \beta = 169 + 2 = 171

Therefore, the correct option is A.

There is a notation discrepancy in the displayed area expression on the solution's, but both extracted approaches conclude the final value as $171$ , and the solution marks Correct Answer: 171.

Common mistakes

Treating the required region as the whole sector instead of the circular segment is incorrect because the area below the line must exclude the triangular part. First identify the chord cut by the line, then use segment area logic.
Using the wrong distance from the center to the line gives an incorrect segment area. For the line $5x - y - 13 = 0$ , use the perpendicular distance formula carefully with coefficients $5, -1, -13$ .
Confusing $\cos^{-1}$ and $\sin^{-1}$ relations leads to sign or angle errors. Use the right-triangle relation consistently when converting between $\cos^{-1}(1/\sqrt{26})$ and an expression involving $\sin^{-1}(12/13)$ .

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Step-by-step solution

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