please-what-does-the-question-mean-by-the-overlapping-portion-of-A-and-B-

Question Number 16592 by chux last updated on 24/Jun/17

please what does the question mean

by the overlapping portion of A

and B .

Commented by chux last updated on 24/Jun/17

Answered by ajfour last updated on 24/Jun/17

Commented by ajfour last updated on 24/Jun/17

Overapping area is the area

common to circles A and B .

Let (h, 0) be the center of circle B .

\sin (x ° / 2) = \frac{h}{2} \Rightarrow h = 2 \sin (x ° / 2)

eqn of circle A :

{(x + h)}^{2} + y^{2} = 1

eqn of circle B :

{(x - h)}^{2} + y^{2} = 1

let circle A intersect x axis at (a, 0) .

then a + h = 1 or a = 1 - h \dots (i)

overlapping area is two times the

area between circle A and y - axis .

upper portion of circle A has eqn .

y = \sqrt{1 - {(x + h)}^{2}}

let required area is S .

S = 4 \int_{0}^{a} \sqrt{1 - {(x + h)}^{2}} dx

= 4 {\frac{x + h}{2} \sqrt{1 - {(x + h)}^{2}} +

\frac{1}{2} \sin^{- 1} (x + h)} ∣_{0}^{a}

= 4 (\frac{a + h}{2}) \sqrt{1 - {(a + h)}^{2}} - \frac{4 h}{2} \sqrt{1 - h^{2}}

+ \frac{4}{2} \sin^{- 1} (a + h) - \frac{4}{2} \sin^{- 1} h

but a + h = 1 and h = 2 \sin (x ° / 2)

S = 0 - 2 h \sqrt{1 - h^{2}} + π - {2 \sin}^{- 1} h

= π - 2 h \sqrt{1 - h^{2}} - {2 \sin}^{- 1} h .

where h = 2 \sin (x ° / 2) .

Commented by chux last updated on 24/Jun/17

thanks alot sir .

Commented by chux last updated on 24/Jun/17

please is it possible to get the

value of x .

Commented by ajfour last updated on 24/Jun/17

that is to be chosen, in question

it said 0 ° ⩽ x ⩽ 60 °, had it said

x = 30 ° or any other value, area

of overlap shall be obtained then

from the derived formula, or if the

area of overlap be given the angle

x can be calculated; they are

interdependent .

Commented by chux last updated on 24/Jun/17

ok sir

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