Question-105422

Question Number 105422 by ajfour last updated on 28/Jul/20

Commented by ajfour last updated on 28/Jul/20

F i n d r a d i u s o f c i r c l e i n t e r m s o f

a a n d b .

Answered by ajfour last updated on 28/Jul/20

l e t A (a - a \cos θ, b + b \sin θ)

s l o p e o f A C = m = - \frac{a \sin θ}{b \cos θ}

\underline{E q u a t i o n s (i) & (i i) :}

x_{C} = a - a \cos θ - r (\frac{b \cos θ}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}})

y_{C} = b + b \sin θ + r (\frac{a \sin θ}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}})

A l s o x_{C} = r \dots . (i i i)

A n d l e t x_{B} = r + r \cos ϕ \dots (i v)

y_{B} = y_{C} + r \sin ϕ \dots . (v)

y_{B} = x_{B}^{2} \dots (v i) \tan ϕ = \frac{1}{2 x_{B}} . . (v i i)

\Rightarrow \frac{1}{4 \tan^{2} ϕ} = y_{C} + r \sin ϕ

\Rightarrow \frac{1}{4 \tan^{2} ϕ} - r \sin ϕ =

b + b \sin θ + r (\frac{a \sin θ}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}})

A n d f r o m (i i i) :

r = a - a \cos θ - r (\frac{b \cos θ}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}})

A n d \frac{1}{2 \tan ϕ} = r + r \cos ϕ

\Rightarrow r = \frac{1}{2 \tan ϕ (1 + \cos ϕ)}

\dots . .

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