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Question Number 164842 by ajfour last updated on 22/Jan/22

Commented by MJS_new last updated on 22/Jan/22

$I think the largest R should be of the triangle$ $with sides 2 a, \sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + b^{2}} \Rightarrow R = \frac{a^{2} + b^{2}}{2 b}$
Commented by ajfour last updated on 22/Jan/22

$I w i s h y o u w e r e w r o n g, s i r . I s h a l l$ $c h e c k . .$
Commented by mr W last updated on 22/Jan/22

$i t h o u g h t t h e s a m e a s M J S s i r a t f i r s t,$ $b e c a u s e i t s e e m s o b v i o u s . b u t t h e n i$ $f o u n d t h i s i s n o t t h e l a r g e s t$ $c i r c u m c i r c l e . i n f a c t f o l l o w i n g a r e$ $t h e l a r g e s t (r e d) a n d s m a l l e s t (g r e e n)$ $c i r c u m c i r c l e s .$
Commented by mr W last updated on 22/Jan/22

Commented by mr W last updated on 22/Jan/22

$R_{m a x} = (\frac{μ}{\sqrt{2} + 1} + \frac{\sqrt{2} + 1}{μ}) \frac{\sqrt{a^{2} + b^{2}}}{4}$ $w i t h μ = \frac{b}{a}$ $e x a m p l e : a = 2, b = 1$ $R_{m a x} = (\frac{\sqrt{2} - 1}{2} + 2 (\sqrt{2} + 1)) \frac{\sqrt{2^{2} + 1^{2}}}{4}$ $R_{m a x} = \frac{3 \sqrt{5} + 5 \sqrt{10}}{8} \approx 2.815$ $\frac{a^{2} + b^{2}}{2 b} = \frac{5}{2} = 2.5 < 2.815$
Commented by MJS_new last updated on 23/Jan/22

${you}^{'} re right . I put a = 1 and use (\begin{matrix} - 1 \\ 0 \end{matrix}) and$ $(\begin{matrix} 0 \\ b \end{matrix}) as vertices of the fixed side if the Δ$ $\Rightarrow the maximum R always is with P = (\begin{matrix} \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} b \end{matrix})$ $and the minimum R with Q = (\begin{matrix} - \frac{\sqrt{2}}{2} \\ - \frac{\sqrt{2}}{2} b \end{matrix})$ $R_{m a x} = \frac{\sqrt{b^{2} + 1}}{4 b} ((\sqrt{2} - 1) b^{2} + \sqrt{2} + 1)$ $R_{m i n} = \frac{\sqrt{b^{2} + 1}}{4 b} ((\sqrt{2} + 1) b^{2} + \sqrt{2} - 1)$ $which represent the same values as you get$
Commented by MJS_new last updated on 23/Jan/22

$the above R_{m i n} is wrong for b below a certain$ $border I^{'} ve not exactly found yet$
Commented by MJS_new last updated on 23/Jan/22

$for b < \sqrt{2} - 1$ $we get 2 points Q_{1}, Q_{2} with x = \frac{b^{2} + 1 \pm \sqrt{b^{4} - 6 b^{2} + 1}}{2 (b^{2} - 1)}$ $and R_{m i n} is below above value$
Commented by MJS_new last updated on 23/Jan/22

$R_{m i n} = \sqrt{- \frac{(b^{2} + 1) ((b^{2} - 1) x - b^{2} - 1) ((b^{2} - 1) \sqrt{1 - x^{2}} + b^{2} + 1)}{8 b^{2}}} with above x$
Commented by mr W last updated on 23/Jan/22

$y o u a r e r i g h t M J S s i r! R_{m a x} i s a l w a y s$ $a t θ = - \frac{π}{4} . b u t R_{m i n} i s a t θ = \frac{3 π}{4} o n l y$ $f o r \frac{b}{a} ⩾ \sqrt{2} - 1 . f o r \frac{b}{a} < \sqrt{2} - 1 t h e r e a r e$ $t w o p o s i t i o n s f o r R_{m i n}, o n e a t θ < \frac{3 π}{4}$ $a n d o n e a t θ > \frac{3 π}{4} . a n d i n t h i s c a s e$ $R_{m i n} = \frac{\sqrt{a^{2} + b^{2}}}{2} .$ $c a n y o u p l e a s e c h e c k i f t h i s m a t c h e s$ $a l s o w i t h y o u r r e s u l t ?$
Commented by mr W last updated on 23/Jan/22

Commented by mr W last updated on 23/Jan/22

Commented by mr W last updated on 23/Jan/22

$s u m m a r y :$ $R_{m a x} = (\frac{μ}{\sqrt{2} + 1} + \frac{\sqrt{2} + 1}{μ}) \frac{\sqrt{a^{2} + b^{2}}}{4}$ $R_{m i n} = (\frac{μ}{\sqrt{2} - 1} + \frac{\sqrt{2} - 1}{μ}) \frac{\sqrt{a^{2} + b^{2}}}{4} f o r μ ⩾ \sqrt{2} - 1$ $R_{m i n} = \frac{\sqrt{a^{2} + b^{2}}}{2} f o r μ < \sqrt{2} - 1$ $w i t h μ = \frac{b}{a} ⩽ 1$
	Answered by ajfour last updated on 22/Jan/22
	$P(acos θ,bsin θ) condider line y−((asin θ)/2)=−(((acos θ+a)/(asin θ)))(x−((a−acos θ)/2)) And line y+(b/2)=(a/b)(x+(a/2)) subteacting (b/2)+((asin θ)/2)=((a/b)+((1+cos θ)/(sin θ)))x +(a^2 /(2b))−((asin θ)/2) ⇒ x=((((a^2 +b^2 )/b)+asin θ)/(2((a/b)+((1+cos θ)/(sin θ))))) y+b=((a^2 +b^2 )/(2b))+(a/b){((((a^2 +b^2 )/b)+asin θ)/(2((a/b)+((1+cos θ)/(sin θ)))))} x^2 +(y+b)^2 =R^2 R^2 ={((((a^2 +b^2 )/b)+asin θ)/(2((a/b)+((1+cos θ)/(sin θ)))))}^2 + +[((a^2 +b^2 )/(2b))+(a/b){((((a^2 +b^2 )/b)+asin θ)/(2((a/b)+((1+cos θ)/(sin θ)))))}]^2 R^2 =(((((a^2 +b^2 )/b)+asin θ)^2 +[4(((a^2 +b^2 )/(2b)))((a/b)+((1+cos θ)/(sin θ)))^2 +(a/b)(((a^2 +b^2 )/b)+asin θ)^2 ]^2 )/(4((a/b)+((1+cos θ)/(sin θ)))^2 )) let for example a=2, b=1 R^2 =(((5+2sin θ)^2 +[10(2+((1+cos θ)/(sin θ)))^2 +2(5+2sin θ)^2 ]^2 )/(4(2+((1+cos θ)/(sin θ)))^2 )) .....$
	$P (a \cos θ, b \sin θ)$ $c o n d i d e r l i n e$ $y - \frac{a \sin θ}{2} = - (\frac{a \cos θ + a}{a \sin θ}) (x - \frac{a - a \cos θ}{2})$ $A n d l i n e$ $y + \frac{b}{2} = \frac{a}{b} (x + \frac{a}{2})$ $s u b t e a c t i n g$ $\frac{b}{2} + \frac{a \sin θ}{2} = (\frac{a}{b} + \frac{1 + \cos θ}{\sin θ}) x$ $+ \frac{a^{2}}{2 b} - \frac{a \sin θ}{2}$ $\Rightarrow x = \frac{\frac{a^{2} + b^{2}}{b} + a \sin θ}{2 (\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}$ $y + b = \frac{a^{2} + b^{2}}{2 b} + \frac{a}{b} {\frac{\frac{a^{2} + b^{2}}{b} + a \sin θ}{2 (\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}}$ $x^{2} + {(y + b)}^{2} = R^{2}$ $R^{2} = {\frac{\frac{a^{2} + b^{2}}{b} + a \sin θ}{2 (\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}}^{2} +$ $+ {[\frac{a^{2} + b^{2}}{2 b} + \frac{a}{b} {\frac{\frac{a^{2} + b^{2}}{b} + a \sin θ}{2 (\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}}]}^{2}$ $R^{2} = \frac{{(\frac{a^{2} + b^{2}}{b} + a \sin θ)}^{2} + {[4 (\frac{a^{2} + b^{2}}{2 b}) {(\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}^{2} + \frac{a}{b} {(\frac{a^{2} + b^{2}}{b} + a \sin θ)}^{2}]}^{2}}{4 {(\frac{a}{b} + \frac{1 + \cos θ}{\sin θ})}^{2}}$ $l e t f o r e x a m p l e a = 2, b = 1$ $R^{2} = \frac{{(5 + 2 \sin θ)}^{2} + {[10 {(2 + \frac{1 + \cos θ}{\sin θ})}^{2} + 2 {(5 + 2 \sin θ)}^{2}]}^{2}}{4 {(2 + \frac{1 + \cos θ}{\sin θ})}^{2}}$ $. . . . .$
	Commented by Tawa11 last updated on 22/Jan/22

	$Great sir$
	Answered by ajfour last updated on 23/Jan/22

	$L e t l e f t e n d b e o r i g i n O .$ $s l o p e p = \frac{b \sin θ}{a + a \cos θ}$ $p^{2} + 1 = \frac{a^{2} {(1 + \cos θ)}^{2} + b^{2} \sin^{2} θ}{a^{2} {(1 + \cos θ)}^{2}}$ $s l o p e q = \frac{b + b \sin θ}{a \cos θ}$ $\tan ϕ = \tan (\tan^{- 1} q - \tan^{- 1} p)$ $R i s m a x i m u m i f ϕ = ϕ_{m a x} .$ $\tan ϕ = \frac{q - p}{1 + p q}$ $\frac{d (\tan ϕ)}{d θ} = 0$ $\Rightarrow \frac{(1 + p q) (\frac{d q}{d θ} - \frac{d p}{d θ}) - (q - p) [q (\frac{d p}{d θ}) + p (\frac{d q}{d θ})]}{{(1 + p q)}^{2}}$ $= 0$ $(1 + p^{2}) (\frac{d q}{d θ}) = (1 + q^{2}) (\frac{d p}{d θ})$ $(1 + p^{2}) [\frac{a b \cos^{2} θ + a \sin θ (b + b \sin θ)}{a^{2} \cos^{2} θ}]$ $= (1 + q^{2}) [\frac{(1 + q^{2}) [a b \cos θ (1 + \cos θ) + a b \sin^{2} θ]}{{(a + a \cos θ)}^{2}}$ $\Rightarrow (1 + p^{2}) (\frac{1 + \sin θ}{\cos^{2} θ}) = (1 + q^{2}) (\frac{1}{1 + \cos θ})$ $. . . .$
	Answered by mr W last updated on 23/Jan/22

	Commented by mr W last updated on 24/Jan/22
	$let μ=(b/a)≤1 A(−a,0) B(0,−b) say P(a cos θ, b sin θ) AP=(√(a^2 (1+cos θ)^2 +b^2 sin^2 θ)) ⇒AP=a(√((1+cos θ)^2 +μ^2 sin^2 θ)) BP=(√(a^2 cos^2 θ+b^2 (1+sin θ)^2 )) ⇒BP=a(√(cos^2 θ+μ^2 (1+sin θ)^2 )) AB=(√(a^2 +b^2 ))=a(√(1+μ^2 )) eqn. of AB: (x/a)+(y/b)+1=0 distance from P to AB: d=((∣cos θ+sin θ+1∣)/( (√((1/a^2 )+(1/b^2 )))))=((a^2 μ∣cos θ+sin θ+1∣)/( (√(a^2 +b^2 )))) area of ΔABP: Δ=(d(√(a^2 +b^2 ))/2)=((a^2 μ(cos θ+sin θ+1))/2) Δ=((AP×BP sin φ)/2) Δ=((a^2 sin φ(√([(1+cos θ)^2 +μ^2 sin^2 θ][cos^2 θ+μ^2 (1+sin θ)^2 ])))/2) ((a^2 μ(cos θ+sin θ+1))/2)=((a^2 sin φ(√([(1+cos θ)^2 +μ^2 sin^2 θ][cos^2 θ+μ^2 (1+sin θ)^2 ])))/2) sin φ=((μ(cos θ+sin θ+1))/( (√([(1+cos θ)^2 +μ^2 sin^2 θ][cos^2 θ+μ^2 (1+sin θ)^2 ])))) R=radius of circumcircle of ΔABP AB=2R sin φ R=((AB)/(2 sin φ)) R=((√(a^2 +b^2 ))/(2μ))×((√([(1+cos θ)^2 +μ^2 sin^2 θ][cos^2 θ+μ^2 (1+sin θ)^2 ]))/(cos θ+sin θ+1)) R=(((1−μ^2 )(√(a^2 +b^2 )))/(2μ))×((√((1+cos θ)(1+sin θ)(((1+μ^2 )/(1−μ^2 ))+cos θ)(((1+μ^2 )/(1−μ^2 ))−sin θ)))/(cos θ+sin θ+1)) let λ=((1+μ^2 )/(1−μ^2 )) R=(((1−μ^2 )(√(a^2 +b^2 )))/(2μ))×((√((1+cos θ)(1+sin θ)(λ+cos θ)(λ−sin θ)))/(cos θ+sin θ+1)) R=(((1−μ^2 )(√(a^2 +b^2 )))/(2μ))×(√(Φ(θ))) Φ(θ)=(((1+cos θ)(1+sin θ)(λ+cos θ)(λ−sin θ))/((cos θ+sin θ+1)^2 )) ln Φ(θ)=ln (1+cos θ)+ln (1+sin θ)+ln (λ+cos θ)+ln (λ−sin θ)−2 ln (cos θ+sin θ+1) ((Φ′(θ))/(Φ(θ)))=((−sin θ)/(1+cos θ))+((cos θ)/(1+sin θ))+((−sin θ)/(λ+cos θ))+((−cos θ)/(λ−sin θ))−((2(−sin θ+cos θ))/(cos θ+sin θ+1)) ((Φ′(θ))/(Φ(θ)))=(cos θ−sin θ)[(((1+cos θ+sin θ)^2 −2(1+cos θ)(1+sin θ))/((1+cos θ)(1+sin θ)))]−(((sin θ+cos θ)(λ+cos θ−sin θ))/((λ+cos θ)(λ−sin θ))) ((Φ′(θ))/(Φ(θ)))=−(((sin θ+cos θ)(λ+cos θ−sin θ))/((λ+cos θ)(λ−sin θ))) Φ′(θ)=0 for R_(max) or R_(min) . (sin θ+cos θ)(λ+cos θ−sin θ)=0 sin θ+cos θ=0 tan θ=−1 ⇒θ=−(π/4) or ((3π)/4) λ+cos θ−sin θ=0 sin θ−cos θ=λ (√2) sin (θ−(π/4))=λ sin (θ−(π/4))=(λ/( (√2))) valid only when (λ/( (√2)))≤1, i.e. λ=((1+μ^2 )/(1−μ^2 ))≤(√2), μ^2 ≤(((√2)−1)/( (√2)+1))=((√2)−1)^2 ⇒μ≤(√2)−1 θ−(π/4)=sin^(−1) (λ/( (√2))) or π−sin^(−1) (λ/( (√2))) ⇒θ=sin^(−1) (λ/( (√2)))+(π/4) or ((5π)/4)−sin^(−1) (λ/( (√2))) R_(max) is at θ=−(π/4): R_(max) =(1/( (√2)))(λ+(1/( (√2))))(((1−μ^2 )(√(a^2 +b^2 )))/(2μ)) R_(max) =((((√2)+1)/( μ))+(μ/( (√2)+1)))((√(a^2 +b^2 ))/4) R_(min) is at θ=((3π)/4), if μ≥(√2)−1: R_(min) =((((√2)−1)/( μ))+(μ/( (√2)−1)))((√(a^2 +b^2 ))/4) R_(min) is at θ=sin^(−1) (λ/( (√2)))+(π/4) or ((5π)/4)−sin^(−1) (λ/( (√2))), if μ≤(√2)−1: R_(min) =((√(a^2 +b^2 ))/2) how to find the circumcenter: eqn. of bisector of AP: y=((b sin θ)/2)−((cos θ+1)/(μ sin θ)) (x−((a(cos θ−1))/2)) eqn. of bisector of BP: y=((b(sin θ−1))/2)−((cos θ)/(μ(1+sin θ)))(x−((a cos θ)/2)) circumcenter C(x_C , y_C ) ((μ(sin θ−1))/2)−((cos θ)/(μ(1+sin θ)))((x_C /a)−((cos θ)/2))=((μ sin θ)/2)−((cos θ+1)/(μ sin θ)) ((x_C /a)−((cos θ−1)/2)) −(μ^2 /2)=((cos θ)/((1+sin θ)))((x_C /a)−((cos θ)/2))−((cos θ+1)/( sin θ)) ((x_C /a)−((cos θ−1)/2)) ((1−μ^2 )/2)=(((cos θ)/(1+sin θ))−((cos θ+1)/( sin θ)))(x_C /a) ⇒(x_C /a)=−(((1−μ^2 )(1+sin θ) sin θ)/(2(1+sin θ+cos θ))) (y_C /a)=((μ(sin θ−1))/2)−((cos θ)/(μ(1+sin θ)))((x_C /a)−((cos θ)/2)) (y_C /a)=(1/(2μ)){((−μ^2 (1−sin θ)(1+sin θ+cos θ)+cos θ−μ^2 sin θ cos θ+cos^2 θ)/((1+sin θ+cos θ)))} ⇒(y_C /a)=(((1−μ^2 )(1+cos θ) cos θ)/(2μ(1+sin θ+cos θ)))$
	$l e t μ = \frac{b}{a} ⩽ 1$ $A (- a, 0)$ $B (0, - b)$ $s a y P (a \cos θ, b \sin θ)$ $A P = \sqrt{a^{2} {(1 + \cos θ)}^{2} + b^{2} \sin^{2} θ}$ $\Rightarrow A P = a \sqrt{{(1 + \cos θ)}^{2} + μ^{2} \sin^{2} θ}$ $B P = \sqrt{a^{2} \cos^{2} θ + b^{2} {(1 + \sin θ)}^{2}}$ $\Rightarrow B P = a \sqrt{\cos^{2} θ + μ^{2} {(1 + \sin θ)}^{2}}$ $A B = \sqrt{a^{2} + b^{2}} = a \sqrt{1 + μ^{2}}$ $e q n . o f A B :$ $\frac{x}{a} + \frac{y}{b} + 1 = 0$ $d i s t a n c e f r o m P t o A B :$ $d = \frac{∣ \cos θ + \sin θ + 1 ∣}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} = \frac{a^{2} μ ∣ \cos θ + \sin θ + 1 ∣}{\sqrt{a^{2} + b^{2}}}$ $a r e a o f Δ A B P :$ $Δ = \frac{d \sqrt{a^{2} + b^{2}}}{2} = \frac{a^{2} μ (\cos θ + \sin θ + 1)}{2}$ $Δ = \frac{A P \times B P \sin ϕ}{2}$ $Δ = \frac{a^{2} \sin ϕ \sqrt{[{(1 + \cos θ)}^{2} + μ^{2} \sin^{2} θ] [\cos^{2} θ + μ^{2} {(1 + \sin θ)}^{2}]}}{2}$ $\frac{a^{2} μ (\cos θ + \sin θ + 1)}{2} = \frac{a^{2} \sin ϕ \sqrt{[{(1 + \cos θ)}^{2} + μ^{2} \sin^{2} θ] [\cos^{2} θ + μ^{2} {(1 + \sin θ)}^{2}]}}{2}$ $\sin ϕ = \frac{μ (\cos θ + \sin θ + 1)}{\sqrt{[{(1 + \cos θ)}^{2} + μ^{2} \sin^{2} θ] [\cos^{2} θ + μ^{2} {(1 + \sin θ)}^{2}]}}$ $R = r a d i u s o f c i r c u m c i r c l e o f Δ A B P$ $A B = 2 R \sin ϕ$ $R = \frac{A B}{2 \sin ϕ}$ $R = \frac{\sqrt{a^{2} + b^{2}}}{2 μ} \times \frac{\sqrt{[{(1 + \cos θ)}^{2} + μ^{2} \sin^{2} θ] [\cos^{2} θ + μ^{2} {(1 + \sin θ)}^{2}]}}{\cos θ + \sin θ + 1}$ $R = \frac{(1 - μ^{2}) \sqrt{a^{2} + b^{2}}}{2 μ} \times \frac{\sqrt{(1 + \cos θ) (1 + \sin θ) (\frac{1 + μ^{2}}{1 - μ^{2}} + \cos θ) (\frac{1 + μ^{2}}{1 - μ^{2}} - \sin θ)}}{\cos θ + \sin θ + 1}$ $l e t λ = \frac{1 + μ^{2}}{1 - μ^{2}}$ $R = \frac{(1 - μ^{2}) \sqrt{a^{2} + b^{2}}}{2 μ} \times \frac{\sqrt{(1 + \cos θ) (1 + \sin θ) (λ + \cos θ) (λ - \sin θ)}}{\cos θ + \sin θ + 1}$ $R = \frac{(1 - μ^{2}) \sqrt{a^{2} + b^{2}}}{2 μ} \times \sqrt{Φ (θ)}$ $Φ (θ) = \frac{(1 + \cos θ) (1 + \sin θ) (λ + \cos θ) (λ - \sin θ)}{{(\cos θ + \sin θ + 1)}^{2}}$ $\ln Φ (θ) = \ln (1 + \cos θ) + \ln (1 + \sin θ) + \ln (λ + \cos θ) + \ln (λ - \sin θ) - 2 \ln (\cos θ + \sin θ + 1)$ $\frac{Φ^{'} (θ)}{Φ (θ)} = \frac{- \sin θ}{1 + \cos θ} + \frac{\cos θ}{1 + \sin θ} + \frac{- \sin θ}{λ + \cos θ} + \frac{- \cos θ}{λ - \sin θ} - \frac{2 (- \sin θ + \cos θ)}{\cos θ + \sin θ + 1}$ $\frac{Φ^{'} (θ)}{Φ (θ)} = (\cos θ - \sin θ) [\frac{{(1 + \cos θ + \sin θ)}^{2} - 2 (1 + \cos θ) (1 + \sin θ)}{(1 + \cos θ) (1 + \sin θ)}] - \frac{(\sin θ + \cos θ) (λ + \cos θ - \sin θ)}{(λ + \cos θ) (λ - \sin θ)}$ $\frac{Φ^{'} (θ)}{Φ (θ)} = - \frac{(\sin θ + \cos θ) (λ + \cos θ - \sin θ)}{(λ + \cos θ) (λ - \sin θ)}$ $Φ^{'} (θ) = 0 f o r R_{m a x} o r R_{m i n} .$ $(\sin θ + \cos θ) (λ + \cos θ - \sin θ) = 0$ $\sin θ + \cos θ = 0$ $\tan θ = - 1$ $\Rightarrow θ = - \frac{π}{4} o r \frac{3 π}{4}$ $λ + \cos θ - \sin θ = 0$ $\sin θ - \cos θ = λ$ $\sqrt{2} \sin (θ - \frac{π}{4}) = λ$ $\sin (θ - \frac{π}{4}) = \frac{λ}{\sqrt{2}}$ $v a l i d o n l y w h e n \frac{λ}{\sqrt{2}} ⩽ 1,$ $i . e . λ = \frac{1 + μ^{2}}{1 - μ^{2}} ⩽ \sqrt{2},$ $μ^{2} ⩽ \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = {(\sqrt{2} - 1)}^{2} \Rightarrow μ ⩽ \sqrt{2} - 1$ $θ - \frac{π}{4} = \sin^{- 1} \frac{λ}{\sqrt{2}} o r π - \sin^{- 1} \frac{λ}{\sqrt{2}}$ $\Rightarrow θ = \sin^{- 1} \frac{λ}{\sqrt{2}} + \frac{π}{4} o r \frac{5 π}{4} - \sin^{- 1} \frac{λ}{\sqrt{2}}$ $R_{m a x} i s a t θ = - \frac{π}{4} :$ $R_{m a x} = \frac{1}{\sqrt{2}} (λ + \frac{1}{\sqrt{2}}) \frac{(1 - μ^{2}) \sqrt{a^{2} + b^{2}}}{2 μ}$ $R_{m a x} = (\frac{\sqrt{2} + 1}{μ} + \frac{μ}{\sqrt{2} + 1}) \frac{\sqrt{a^{2} + b^{2}}}{4}$ $R_{m i n} i s a t θ = \frac{3 π}{4}, i f μ ⩾ \sqrt{2} - 1 :$ $R_{m i n} = (\frac{\sqrt{2} - 1}{μ} + \frac{μ}{\sqrt{2} - 1}) \frac{\sqrt{a^{2} + b^{2}}}{4}$ $R_{m i n} i s a t θ = \sin^{- 1} \frac{λ}{\sqrt{2}} + \frac{π}{4} o r \frac{5 π}{4} - \sin^{- 1} \frac{λ}{\sqrt{2}}, i f μ ⩽ \sqrt{2} - 1 :$ $R_{m i n} = \frac{\sqrt{a^{2} + b^{2}}}{2}$ $h o w t o f i n d t h e c i r c u m c e n t e r :$ $e q n . o f b i s e c t o r o f A P :$ $y = \frac{b \sin θ}{2} - \frac{\cos θ + 1}{μ \sin θ} (x - \frac{a (\cos θ - 1)}{2})$ $e q n . o f b i s e c t o r o f B P :$ $y = \frac{b (\sin θ - 1)}{2} - \frac{\cos θ}{μ (1 + \sin θ)} (x - \frac{a \cos θ}{2})$ $c i r c u m c e n t e r C (x_{C}, y_{C})$ $\frac{μ (\sin θ - 1)}{2} - \frac{\cos θ}{μ (1 + \sin θ)} (\frac{x_{C}}{a} - \frac{\cos θ}{2}) = \frac{μ \sin θ}{2} - \frac{\cos θ + 1}{μ \sin θ} (\frac{x_{C}}{a} - \frac{\cos θ - 1}{2})$ $- \frac{μ^{2}}{2} = \frac{\cos θ}{(1 + \sin θ)} (\frac{x_{C}}{a} - \frac{\cos θ}{2}) - \frac{\cos θ + 1}{\sin θ} (\frac{x_{C}}{a} - \frac{\cos θ - 1}{2})$ $\frac{1 - μ^{2}}{2} = (\frac{\cos θ}{1 + \sin θ} - \frac{\cos θ + 1}{\sin θ}) \frac{x_{C}}{a}$ $\Rightarrow \frac{x_{C}}{a} = - \frac{(1 - μ^{2}) (1 + \sin θ) \sin θ}{2 (1 + \sin θ + \cos θ)}$ $\frac{y_{C}}{a} = \frac{μ (\sin θ - 1)}{2} - \frac{\cos θ}{μ (1 + \sin θ)} (\frac{x_{C}}{a} - \frac{\cos θ}{2})$ $\frac{y_{C}}{a} = \frac{1}{2 μ} {\frac{- μ^{2} (1 - \sin θ) (1 + \sin θ + \cos θ) + \cos θ - μ^{2} \sin θ \cos θ + \cos^{2} θ}{(1 + \sin θ + \cos θ)}}$ $\Rightarrow \frac{y_{C}}{a} = \frac{(1 - μ^{2}) (1 + \cos θ) \cos θ}{2 μ (1 + \sin θ + \cos θ)}$
	Commented by mr W last updated on 23/Jan/22

	Commented by mr W last updated on 23/Jan/22

	Commented by mr W last updated on 23/Jan/22

	Commented by ajfour last updated on 24/Jan/22

	$T h a n k s S i r, g r e a t e f f o r t a n d$ $s u c c e s f u l w o r k i n g s, i^{'} l l g o$ $t h r o u g h v e r y s o o n . .$
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