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Question Number 46889 by ajfour last updated on 02/Nov/18

Commented by ajfour last updated on 02/Nov/18

$Q . 46611 (A l t e r n a t e a p p r o a c h)$
	Answered by ajfour last updated on 02/Nov/18

	$\hat{n} = \cos θ (\sin α \hat{i} + \cos α \hat{j}) + \sin θ \hat{k}$ ${\bar{r}}_{M} = (h \hat{i} + k \hat{j}) - R \sin θ (\sin α \hat{i} + \cos α \hat{j})$ $+ R \cos θ \hat{k}$ $a \sin α = d$ $b \cos α = d$ $c \tan θ = d$ $y_{P} = 0$ $& {[\frac{\hat{n} \times (c \hat{k} - a \hat{i})}{\sqrt{c^{2} + a^{2}}}]}_{y} = {({\bar{r}}_{P} - {\bar{r}}_{M})}_{y}$ $\Rightarrow - R [\frac{c \cos θ \sin α + a \sin θ}{\sqrt{a^{2} + c^{2}}}]$ $= - k + R \sin θ \cos α . . . (i)$ $x_{Q} = 0 &$ ${[\frac{(c \hat{k} - b \hat{j}) \times \hat{n}}{\sqrt{b^{2} + c^{2}}}]}_{x} = {({\bar{r}}_{Q} - {\bar{r}}_{M})}_{x}$ $\Rightarrow - R [\frac{c \cos θ \cos α + b \sin θ}{\sqrt{b^{2} + c^{2}}}]$ $= - h + R \sin θ \sin α . . . . (i i)$ $F r o m (i) & (i i), w e s h a l l o b t a i n$ $α a n d θ; u s i n g$ $a \sin α = d$ $b \cos α = d$ $c \tan θ = d, i n e q s . (i) & (i i)$ $\frac{R [\frac{\cos^{2} θ}{\sin θ} \sin α + \frac{\sin θ}{\sin α}]}{\sqrt{\frac{1}{\sin^{2} α} + \frac{\cos^{2} θ}{\sin^{2} θ}}} = k - R \sin θ \cos α$ $s i m i l a r l y$ $\frac{R [\frac{\cos^{2} θ}{\sin θ} \cos α + \frac{\sin θ}{\cos α}]}{\sqrt{\frac{1}{\cos^{2} α} + \frac{\cos^{2} θ}{\sin^{2} θ}}} = h - R \sin θ \sin α$ $________________________$ $r e a r r a n g i n g g i v e s$ $R \sqrt{\sin^{2} θ + \sin^{2} α \cos^{2} θ}$ $= k - R \sin θ \cos α . . . (A)$ $R \sqrt{\sin^{2} θ + \cos^{2} θ \cos^{2} α}$ $= h - R \sin θ \sin α . . . . (B)$ $________________________$ $s q u a r i n g a n d a d d i n g (A) & (B)$ $2 R^{2} \sin^{2} θ + R^{2} \cos^{2} θ = h^{2} + k^{2} +$ $R^{2} \sin^{2} θ - 2 R \sin θ (h \sin α + k \cos α)$ $\Rightarrow 2 R \sin θ = \frac{h^{2} + k^{2} - R^{2}}{h \sin α + k \cos α} . . (I)$ $C h a n g i n g (A) t o$ $R^{2} \sin^{2} θ \cos^{2} α + R^{2} \sin^{2} α =$ ${(k - R \sin θ \cos α)}^{2}$ $N o w l e t R \sin θ \cos α = t$ $\Rightarrow t^{2} + R^{2} \sin^{2} α = k^{2} + t^{2} - 2 k t$ $\Rightarrow 2 t = \frac{k^{2} - R^{2} \sin^{2} α}{k}$ $N o w u s i n g (I) :$ $\Rightarrow 2 R \sin θ = \frac{k}{\cos α} - \frac{R^{2} \sin^{2} α}{k \cos α}$ $\Rightarrow \frac{h^{2} + k^{2} - R^{2}}{h \sin α + k \cos α} = \frac{k^{2} - R^{2} \sin^{2} α}{k \cos α}$ $\Rightarrow \frac{h^{2} - R^{2} \cos^{2} α}{h \sin α} = \frac{k^{2} - R^{2} \sin^{2} α}{k \cos α}$ $________________________$ $l e t m = \tan α$ $\Rightarrow \frac{h^{2} (1 + m^{2}) - R^{2}}{k^{2} (1 + m^{2}) - R^{2} m^{2}} = \frac{m h}{k}$ $t h i s e q . p r o v i d e s f o r α; a n d t h e n$ $f r o m (I) :$ $θ = \sin^{- 1} [\frac{h^{2} + k^{2} - R^{2}}{2 R (h \sin α + k \cos α)}]$ $________________________.$
	Commented by ajfour last updated on 02/Nov/18

	$I n e x c e l l e n t a g r e e m e n t w i t h$ $y o u r r e s u l t S i r, c o z t h e y a r e s a m e .$
	Commented by MrW3 last updated on 02/Nov/18

	$y e s, s i r .$ ${i t}^{'} s a m a r v e l l o u s j o b w h a t y o u d i d!$ $r e a l l y n i c e m e t h o d!$
	Commented by MrW3 last updated on 02/Nov/18

	$Y o u c a n r e m e b e r t h a t I w a s u n s u r e i f$ $t h e p o s i t i o n o f t h e r i n g i s u n i q u e . N o w$ $I k n o w t h e p o s i t i o n o f t h e r i n g i s$ $u n i q u e w i t h t h e g i v e n p o i n t G . B u t$ $w i t h s o m e v a l u e s o f R t h e r e c a n b e$ $t w o o r t h r e e p o s s i b l e p o s i t i o n s f o r$ $t h e r i n g .$
	Commented by ajfour last updated on 02/Nov/18

	$y e s s i r, t h a n k y o u f o r s o l v i n g i t$ $b y t w o m e t h o d s; b u t i c o u l d$ $h a v e g u e s s e d w r o n g l y a t l e a s t$ $t w o r o o t s f o r α a s c o m p l e x . .$ $e v e r y t i m e .$
	Commented by MrW3 last updated on 02/Nov/18

	$t h e r e c o u l d b e t h r e e p o s i t i v e r o o t s o r$ $o n e p o s i t i v e a n d t w o n e g a t i v e o r t w o$ $p o s i t i v e a n d o n e n e g a t i v e r o o t s . t h a t$ $m e a n s w e c a n g e t : o n e p o s i t i o n, t w o$ $p o s i t i o n s o r t h r e e p o s i t i o n s f o r t h e$ $r i n g .$
	Commented by peter frank last updated on 02/Nov/18

	$pls help Q 46838, 46829$
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