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Question Number 105001 by ajfour last updated on 25/Jul/20

Commented by ajfour last updated on 25/Jul/20

$I f b o t h c i r c l e s h a v e u n i t r a d i u s, a n d$ $r e g i o n s 1, 2, 3, 4, 5 h a v e e q u a l a r e a s,$ $f i n d e q . o f b o t h c i r c l e s .$
	Answered by ajfour last updated on 25/Jul/20

	$l e t l o w e r c i r c l e e q . b e$ ${(x - h)}^{2} + {(y + k)}^{2} = 1$ $u p p e r c i r c l e e q . i s t h e n$ ${(x + k)}^{2} + {(y - h)}^{2} = 1$ $l e t A (a, 0) & B (b, 0)$ $l e t x_{0} = a, b$ $\Rightarrow x_{0} = h \pm \sqrt{1 - k^{2}}$ $\Rightarrow a = h + \sqrt{1 - k^{2}}, b = h - \sqrt{1 - k^{2}}$ $P (p, p) l i e s o n y = x a n d b o t h c i r c l e s$ $h e n c e {(p + k)}^{2} + {(p - h)}^{2} = 1$ $A_{5} = \int_{b}^{0} (- k + \sqrt{1 - {(x - h)}^{2}}) d x$ $A_{2} + A_{3} = \int_{0}^{a} (- k + \sqrt{1 - {(x - h)}^{2}}) d x$ $A_{3} = 2 \int_{0}^{p} (- k + \sqrt{1 - {(x - h)}^{2}} - x) d x$ $N o w A_{2} + A_{3} = 2 A_{5} . . . . (i) &$ $A_{3} = A_{5} . . . . (i i)$ $. . . . .$
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