Line $x = 1$ divides $A$ enclosed by circle $x^{2} + y^{2} = 16$ in two portions $A_{1}$ and $A_{2}(A_{1} A_{2})$, then $\frac{A_{1}}{A_{2}}$ is

Question

Line $x = 1$ divides $A$ enclosed by circle $x^{2} + y^{2} = 16$ in two portions $A_{1}$ and $A_{2}(A_{1} > A_{2})$, then $\frac{A_{1}}{A_{2}}$ is

Dr. Abubacker · Accepted Answer

(d)
Equation of circle is $x^{2} + y^{2} = 16$
$	herefore$ Total area of circle $= A_{1} + A_{2} = 16\pi$ …(i)
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$\frac{A_{1}}{A_{2}} = \frac{16\pi}{A_{2}} - 1$ [on dividing Eq. (i) by $A_{2}$]
and $A_{2} = 2\int_{1}^{4}{\sqrt{16 - x^{2}}dx}$
$A_{2} = 2\left\{ \frac{x}{2}\sqrt{16 - x^{2}} + \frac{16}{2}\sin^{- 1}\left( \frac{x}{4} ight) ight\}_{1}^{4}$
$= 2\left\{ 4\pi - \frac{\sqrt{15}}{2} - 8\sin^{- 1}\left( \frac{1}{4} ight) ight\}$
$= 8\pi - \sqrt{15} - 16\sin^{- 1}\left( \frac{1}{4} ight)$
$	herefore\ \frac{A_{1}}{A_{2}} = \frac{16\pi}{8\pi - \sqrt{15} - 16\sin^{- 1}\left( \frac{1}{4} ight)} - 1$

Q. Line $x = 1$ divides $A$ enclosed by circle $x^{2} + y^{2} = 16$ in two portions $A_{1}$ and $A_{2}(A_{1} > A_{2})$, then $\frac{A_{1}}{A_{2}}$ is

Explanation

Reviewed by Dr. Abubacker

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