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Question Number 46935 by ajfour last updated on 02/Nov/18

Commented by ajfour last updated on 03/Nov/18

$I f b o t h c o l o u r e d r e g i o n s h a v e t h e$ $s a m e a r e a, f i n d l o c u s o f C i n$ $p o l a r f o r m o r c a r t e s i a n f o r m .$ $T a k e c e n t e r o f r e c t a n g l e a s o r i g i n,$ $a n d r a d i u s o f c i r c l e R .$ $r = \sqrt{x_{C}^{2} + y_{C}^{2}}, \tan θ = \frac{y_{C}}{x_{C}} .$
	Answered by ajfour last updated on 04/Nov/18

	$l e t c e n t e r o f c i r c l e b e o r i g i n .$ $E q . i s x = R \cos ϕ; y = R \sin ϕ$ $c e n t e r o f r e c t a n g l e (- r \cos θ, - r \sin θ)$ $ϕ = π + θ . . . (i)$ $u p p e r e d g e : y = - r \sin θ + \frac{b}{2}$ $r i g h t v e r t i c a l e d g e :$ $x = \frac{a}{2} - r \cos θ$ $L e t p o i n t s o f i n t e r s e c t i o n b e$ $A (l e f t), B (r i g h t) .$ $\Rightarrow R \sin ϕ_{A} = - r \sin θ_{A} + \frac{b}{2}$ $u s i n g (i)$ $\Rightarrow R \sin θ_{A} = r \sin θ_{A} - \frac{b}{2}$ $\Rightarrow \sin θ_{A} = \frac{b}{2 (r - R)}$ $________________________$ $R \cos ϕ_{B} = \frac{a}{2} - r \cos θ_{B}$ $\Rightarrow R \cos θ_{B} = r \cos θ_{B} - \frac{a}{2}$ $\cos θ_{B} = \frac{a}{2 (r - R)}$ $. . . .$
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