Application of Integrals

The area (in sq. units) of the part of the circle x2+y2=169 which is below the line 5x-y=13 is 2 - 652+ -1 ( 1213 ), where , are coprime numbers. Then + is equal to .

Answer

171

Solution

The line meets the circle where x2 + (5x - 13)2 = 169, i.e. 26x(x - 5) = 0, so at A(5, 12) and B(0, − 13).

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Below the line means y < 5x - 13, i.e. x > y + 135, so for each y from − 13 to 12 a horizontal strip runs from the line x1 = y + 135 to the arc x2 = 169 - y2.

Hence Area = -1312 ( 169 - y2 - y + 135 )dy

-1312 169 - y2 dy = [ y2 169 - y2 + 1692 -1 y13 ]-1312 = 30 + 1692 -1 1213 + 169 4

-1312 y + 135 dy = 110 [(y + 13)2 ]-1312 = 1252

So Area = 169 4 - 652 + 1692 -1 ( 1213 )

Comparing with the given form, = 1692, i.e. α = 169, β = 2.

∴ α + β = 171

Answer: 171