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Question Number 44801 by ajfour last updated on 04/Oct/18

Commented by ajfour last updated on 04/Oct/18

$E q u a t i o n o f e l l i p s e : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ $f i n d r a d i u s r o f c i r c l e .$
	Answered by ajfour last updated on 05/Oct/18
	$let centre of circle be (h,0) h = (r/(sin tan^(−1) m)) = ((r(√(1+m^2 )))/m) point of contact (upper) : (x_0 ,y_0 ) { ((x_0 =acos φ = h+rcos θ )),((y_0 = bsin φ = rsin θ)) :} tan φ = ((arsin θ)/(b(h+rcos θ))) ....(i) (dy/dx)∣_x_0 = ((bcos φ)/(−asin φ)) = −((cos θ)/(sin θ)) ⇒ tan φ = ((bsin θ)/(acos θ)) ...(ii) comparing (i)&(ii) ((arsin θ)/(b(h+rcos θ))) = ((bsin θ)/(acos θ)) as θ≠0 , a^2 rcos θ = b^2 h+b^2 rcos θ ________________________ { ((if θ = 0 ⇒ a = b, then)),(( r = a−h = a−((r(√(1+m^2 )))/m))) :} ⇒ r = (a/(1+((√(1+m^2 ))/m))) ________________________ But when its not so ⇒ cos θ = ((b^2 h)/(r(a^2 −b^2 ))) from first bracket eqs. (((h+rcos θ)^2 )/a^2 ) +(((rsin θ)^2 )/b^2 ) = 1 ⇒ b^2 (h+rcos θ)^2 +a^2 (r^2 −r^2 cos^2 θ) = a^2 b^2 b^2 (h+((b^2 h)/(a^2 −b^2 )))^2 +a^2 [r^2 −(((b^2 h)/(a^2 −b^2 )))^2 ] = a^2 b^2 ⇒ r^2 = ((b^4 h^2 )/((a^2 −b^2 )^2 ))+b^2 −((a^2 b^2 h^2 )/((a^2 −b^2 )^2 )) r^2 = b^2 (1−(h^2 /(a^2 −b^2 ))) but h^2 = r^2 +(r^2 /m^2 ) ⇒ r^2 = b^2 −((b^2 r^2 )/(a^2 −b^2 ))−((b^2 r^2 )/(m^2 (a^2 −b^2 ))) r^2 [1+((b^2 /(a^2 −b^2 )))(1+(1/m^2 ))]= b^2 ________________________ r = (b/(√(1+((b^2 /(a^2 −b^2 )))(1+(1/m^2 ))))) If θ ≠ 0 (same as a ≠ b).$
	$l e t c e n t r e o f c i r c l e b e (h, 0)$ $h = \frac{r}{\sin \tan^{- 1} m} = \frac{r \sqrt{1 + m^{2}}}{m}$ $p o i n t o f c o n t a c t (u p p e r) : (x_{0}, y_{0})$ ${\begin{cases} x_{0} = a \cos ϕ = h + r \cos θ \\ y_{0} = b \sin ϕ = r \sin θ \end{cases}$ $\tan ϕ = \frac{a r \sin θ}{b (h + r \cos θ)} . . . . (i)$ $\frac{d y}{d x} ∣_{x_{0}} = \frac{b \cos ϕ}{- a \sin ϕ} = - \frac{\cos θ}{\sin θ}$ $\Rightarrow \tan ϕ = \frac{b \sin θ}{a \cos θ} . . . (i i)$ $c o m p a r i n g (i) & (i i)$ $\frac{a r \sin θ}{b (h + r \cos θ)} = \frac{b \sin θ}{a \cos θ}$ $a s θ \neq 0, a^{2} r \cos θ = b^{2} h + b^{2} r \cos θ$ $________________________$ ${\begin{cases} i f θ = 0 \Rightarrow a = b, t h e n \\ r = a - h = a - \frac{r \sqrt{1 + m^{2}}}{m} \end{cases}$ $\Rightarrow r = \frac{a}{1 + \frac{\sqrt{1 + m^{2}}}{m}}$ $________________________$ $B u t w h e n i t s n o t s o$ $\Rightarrow \cos θ = \frac{b^{2} h}{r (a^{2} - b^{2})}$ $f r o m f i r s t b r a c k e t e q s .$ $\frac{{(h + r \cos θ)}^{2}}{a^{2}} + \frac{{(r \sin θ)}^{2}}{b^{2}} = 1$ $\Rightarrow b^{2} {(h + r \cos θ)}^{2} + a^{2} (r^{2} - r^{2} \cos^{2} θ)$ $= a^{2} b^{2}$ $b^{2} {(h + \frac{b^{2} h}{a^{2} - b^{2}})}^{2} + a^{2} [r^{2} - {(\frac{b^{2} h}{a^{2} - b^{2}})}^{2}]$ $= a^{2} b^{2}$ $\Rightarrow r^{2} = \frac{b^{4} h^{2}}{{(a^{2} - b^{2})}^{2}} + b^{2} - \frac{a^{2} b^{2} h^{2}}{{(a^{2} - b^{2})}^{2}}$ $r^{2} = b^{2} (1 - \frac{h^{2}}{a^{2} - b^{2}})$ $b u t h^{2} = r^{2} + \frac{r^{2}}{m^{2}}$ $\Rightarrow r^{2} = b^{2} - \frac{b^{2} r^{2}}{a^{2} - b^{2}} - \frac{b^{2} r^{2}}{m^{2} (a^{2} - b^{2})}$ $r^{2} [1 + (\frac{b^{2}}{a^{2} - b^{2}}) (1 + \frac{1}{m^{2}})] = b^{2}$ $________________________$ $r = \frac{b}{\sqrt{1 + (\frac{b^{2}}{a^{2} - b^{2}}) (1 + \frac{1}{m^{2}})}}$ $I f θ \neq 0 (s a m e a s a \neq b) .$
	Commented by ajfour last updated on 05/Oct/18

	$I f a = 5, b = 4, m = 1$ $r = \frac{4}{\sqrt{1 + \frac{32}{9}}} = \frac{12}{\sqrt{41}} .$
	Answered by MrW3 last updated on 05/Oct/18

	$e q n . o f e l l i p s e :$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} y^{'} = 0$ $\Rightarrow y^{'} = - \frac{b^{2} x}{a^{2} y}$ $c e n t e r o f c i r c l e (x_{0}, 0) w i t h$ $x_{0} = \frac{\sqrt{1 + m^{2}}}{m} r = r \sqrt{1 + \frac{1}{m^{2}}} = μ r$ $c o n t a c t p o i n t (u, v)$ $u = x_{0} + r \cos θ = r (μ + \cos θ)$ $v = r \sin θ$ $\frac{r^{2} {(μ + \cos θ)}^{2}}{a^{2}} + \frac{r^{2} \sin^{2} θ}{b^{2}} = 1$ $\frac{1}{\tan θ} = \frac{b^{2} u}{a^{2} v} = \frac{b^{2} r (μ + \cos θ)}{a^{2} r \sin θ} = \frac{b^{2} (μ + \cos θ)}{a^{2} \sin θ}$ $a^{2} \cos θ = b^{2} (μ + \cos θ)$ $(a^{2} - b^{2}) \cos θ = μ b^{2}$ $\cos θ = \frac{μ b^{2}}{a^{2} - b^{2}} = \frac{μ}{λ^{2} - 1} w i t h λ = \frac{a}{b}$ $r^{2} [\frac{{(μ + \cos θ)}^{2}}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}] = 1$ $r^{2} [\frac{b^{2} {(μ + \cos θ)}^{2} + a^{2} \sin^{2} θ}{a^{2} b^{2}}] = 1$ $r^{2} [\frac{{(μ + \frac{μ}{λ^{2} - 1})}^{2} + λ^{2} (1 - \frac{μ}{λ^{2} - 1}) (1 + \frac{μ}{λ^{2} - 1})}{a^{2}}] = 1$ $r^{2} [\frac{\frac{μ^{2} λ^{2} (λ^{2} - 1) + λ^{2} {(λ^{2} - 1)}^{2}}{{(λ^{2} - 1)}^{2}}}{a^{2}}] = 1$ $r^{2} [\frac{λ^{2} (μ^{2} + λ^{2} - 1)}{a^{2} (λ^{2} - 1)}] = 1$ $\Rightarrow r = \frac{a}{λ} \sqrt{\frac{λ^{2} - 1}{μ^{2} + λ^{2} - 1}}$ $\Rightarrow r = b \sqrt{\frac{λ^{2} - 1}{μ^{2} + λ^{2} - 1}}$ $\Rightarrow r = b \sqrt{\frac{{(\frac{a}{b})}^{2} - 1}{1 + \frac{1}{m^{2}} + {(\frac{a}{b})}^{2} - 1}}$ $\Rightarrow r = b \sqrt{\frac{{(\frac{a}{b})}^{2} - 1}{\frac{1}{m^{2}} + {(\frac{a}{b})}^{2}}}$ $r = \frac{b}{\sqrt{1 + (1 + \frac{1}{m^{2}}) (\frac{b^{2}}{a^{2} - b^{2}})}}$
	Commented by ajfour last updated on 05/Oct/18

	$T h a n k y o u S i r, i h o p e m y$ $a n s w e r m a t c h e s . .$
	Commented by MrW3 last updated on 05/Oct/18

	$o u r a n s w e r s a r e t h e s a m e .$
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