Question-216060

Question Number 216060 by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

f i n d r a d i u s o f i n s c r i b e d c i r c l e r = ?

Answered by mr W last updated on 26/Jan/25

s a y μ = \frac{b}{a}, λ = \frac{r}{a}

s a y P (a \cos θ, b \sin θ)

\tan ϕ = - \frac{d x}{d y} = \frac{a \sin θ}{b \cos θ} = \frac{\tan θ}{μ} = \frac{m}{μ}

b \sin θ = r \sin ϕ

\Rightarrow μ \sin θ = \frac{λ m}{\sqrt{m^{2} + μ^{2}}} \dots (i)

a \cos θ = r + r \cos ϕ

\Rightarrow \cos θ = λ (1 + \frac{μ}{\sqrt{m^{2} + μ^{2}}}) \dots (i i)

(i) / (i i) :

μ \tan θ = \frac{m}{μ + \sqrt{m^{2} + μ^{2}}}

μ = \frac{1}{μ + \sqrt{m^{2} + μ^{2}}}

\sqrt{m^{2} + μ^{2}} = \frac{1}{μ} - μ

\Rightarrow m^{2} = \frac{1}{μ^{2}} - 2 ⩾ 0 \Rightarrow μ ⩽ \frac{1}{\sqrt{2}} \Rightarrow a ⩾ \sqrt{2} b

f r o m (i) :

μ \sin θ = \frac{λ m}{\sqrt{m^{2} + μ^{2}}}

\frac{μ m}{\sqrt{1 + m^{2}}} = \frac{λ m}{\sqrt{m^{2} + μ^{2}}}

\Rightarrow λ = \frac{μ \sqrt{m^{2} + μ^{2}}}{\sqrt{1 + m^{2}}} = \frac{μ (\frac{1}{μ} - μ)}{\sqrt{\frac{1}{μ^{2}} - 1}} = μ \sqrt{1 - μ^{2}}

i . e . r = b \sqrt{1 - {(\frac{b}{a})}^{2}}

Commented by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

Commented by ajfour last updated on 27/Jan/25

I f a ⩽ \sqrt{2} b t h e n s u c h r = \frac{a}{2}

A n d i f a ⩾ \sqrt{2} b r = \frac{b}{a} \sqrt{a^{2} - b^{2}}

I s n t i t t h i s t o s u m m a r i z e, S i r ? T h a n k s .

Commented by mr W last updated on 27/Jan/25

i f a = \sqrt{2} b t h e n r = \frac{a}{2} .

i f a < \sqrt{2} b t h e n n o i n s c r i b e d c i r c l e

p o s s i b l e .

Answered by aleks041103 last updated on 27/Jan/25

Commented by aleks041103 last updated on 27/Jan/25

T h e c i r c l e i s t a n g e n t t o t h e e l l i p s e a n d t h e r e f o r e

t h e r a d i u s i s ⊥ t o t h e t a n g e n t t o t h e e l l i p s e

a t t h i s p o i n t .

T h e t a n g e n t o f a n e l l i p s e i s ⊥ t o t h e a n g l e

b i s e c t o r o f t h e a n g l e b e t w e e n t h e s e g m e n t s

c o n n e c t i n g p o i n t t h e f o c i t o t h e p o i n t .

T h u s w e g e t t h e t o p s c h e m a t i c .

N o w :

x + y = 2 a \Rightarrow y = 2 a - x

\frac{x}{y} = \frac{c + r}{c - r}

x y - (c + r) (c - r) = r^{2} \Rightarrow x y = c^{2} = a^{2} - b^{2}

c = a \sqrt{1 - {(\frac{b}{a})}^{2}}

\Rightarrow x (2 a - x) = a^{2} - b^{2}

\Rightarrow {(x - a)}^{2} - b^{2} = 0

\Rightarrow x = a + b; a - b

b u t x > y \Rightarrow x = a + b, y = a - b

\Rightarrow \frac{x}{h} = \frac{a + b}{a - b} = \frac{c + r}{c - r}

\Rightarrow r = \frac{b c}{a}

\Rightarrow r = b \sqrt{1 - {(\frac{b}{a})}^{2}}

Commented by aleks041103 last updated on 27/Jan/25

T h e r a d i u s t o t h e p o i n t P (o r i g i n a l s c h e m a t i c)

i s p e r p e n d i c u l a r (⊥) t o t h e t a n g e n t o f t h e

e l l i p s e a t P (s i n c e t h e t a n g e n t o f t h e c i r c l e

c o i n s i d e s w i t h t h e t a n g e n t t o t h e e l l i p s e) .

Commented by mr W last updated on 27/Jan/25

t h a n k s f o r t h i s n i c e s o l u t i o n!

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