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Question Number 197128 by Mingma last updated on 08/Sep/23

Commented by Frix last updated on 09/Sep/23

$I get ρ \approx . 213841779235$
	Answered by sniper237 last updated on 09/Sep/23
	$The circle is embedded in the triangle formed by the intersection points of curves y=(√(x )) ; y=(√(1−x^2 )) and y=0 They are A(0,0), B(0,1) ,C(ϕ^(−1) ,(√ϕ) ). Let O, r, the center and radius of the circle For each triangle OAB,OAC,OBC the morgue is r. So if S is ABC area ((r×AB)/2)+((r×AC)/2)+((r×BC)/2)=S then r = ((2S)/(AB+AC+BC)) ABC=D={ (x,y)/ (√x) ≤y≤(√(1−x^2 )) } So S= ∫∫_D dxdy=∫_0 ^ϕ^(−1) ((√(1−x^2 )) −(√x) )dx$
	$T h e c i r c l e i s e m b e d d e d i n t h e t r i a n g l e$ $f o r m e d b y t h e i n t e r s e c t i o n p o i n t s o f$ $c u r v e s y = \sqrt{x}; y = \sqrt{1 - x^{2}} a n d y = 0$ $T h e y a r e A (0, 0), B (0, 1), C (φ^{- 1}, \sqrt{φ}) . L e t$ $O, r, t h e c e n t e r a n d r a d i u s o f t h e c i r c l e$ $F o r e a c h t r i a n g l e O A B, O A C, O B C$ $t h e m o r g u e i s r . S o i f S i s A B C a r e a$ $\frac{r \times A B}{2} + \frac{r \times A C}{2} + \frac{r \times B C}{2} = S$ $t h e n r = \frac{2 S}{A B + A C + B C}$ $A B C = D = {(x, y) / \sqrt{x} ⩽ y ⩽ \sqrt{1 - x^{2}}}$ $S o S = \int \int_{D} d x d y = \int_{0}^{φ^{- 1}} (\sqrt{1 - x^{2}} - \sqrt{x}) d x$
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