Question-101375

Question Number 101375 by ajfour last updated on 02/Jul/20

Commented by ajfour last updated on 02/Jul/20

I n t e r m s o f R, f i n d r a d i i a, b, c .

Commented by ajfour last updated on 05/Jul/20

Commented by ajfour last updated on 06/Jul/20

l e t s c o n s i d e r b o t h c i r c l e s o f r a d i i

a a n d c g e n e r a l l y .

B o t h a r e a t a d i s t a n c e o f (R + r) f r o m

c e n t e r M o f c i r c l e o f r a d i u s R .

B o t h t o u c h t h e l i n e y = R^{2} + \frac{1}{4} + R

b o t h t o u c h t h e p a r a b o l a a t (x, x^{2})

x_{1} = \sqrt{R^{2} - \frac{1}{4}}

c e n t e r o f o u r g e n e r a l c i r c l e i s

(2 \sqrt{R r}, R^{2} + \frac{1}{4} + R - r)

e q . o f t h e c i r c l e :

{\begin{cases} {(x - 2 \sqrt{R r})}^{2} + {[x^{2} - (R^{2} + \frac{1}{4} + R - r)]}^{2} = r^{2} \\ {(R + \frac{1}{2})}^{2} - r - \frac{r}{\sqrt{1 + 4 x^{2}}} = x^{2} \end{cases}

a n d x - 2 \sqrt{R r} = \frac{r (2 x)}{\sqrt{1 + 4 x^{2}}} f o r r = c

b u t 2 \sqrt{R r} - x = \frac{r (2 x)}{\sqrt{1 + 4 x^{2}}} f o r r = a

A n d s i n c e w e k n o w x_{1} = \sqrt{R^{2} - \frac{1}{4}}

2 \sqrt{{R r}_{1}} = x + \frac{2 x_{1} r}{\sqrt{1 + 4 x_{1}^{2}}}

2 \sqrt{R a} = \sqrt{R^{2} - \frac{1}{4}} + \frac{2 a \sqrt{R^{2} - \frac{1}{4}}}{\sqrt{1 + 4 R^{2} - 1}}

2 \sqrt{R a} = \sqrt{R^{2} - \frac{1}{4}} + \frac{a}{R} \sqrt{R^{2} - \frac{1}{4}}

\frac{a}{R} \sqrt{R^{2} - \frac{1}{4}} - 2 R \sqrt{\frac{a}{R}} + \sqrt{R^{2} - \frac{1}{4}} = 0

\sqrt{\frac{a}{R}} = \frac{R}{\sqrt{R^{2} - \frac{1}{4}}} \pm \sqrt{\frac{R^{2}}{R^{2} - \frac{1}{4}} - 1}

\sqrt{\frac{a}{R}} = \frac{R \pm \frac{1}{2}}{\sqrt{R^{2} - \frac{1}{4}}}

\Rightarrow \frac{a}{R} = \frac{{(R \pm \frac{1}{2})}^{2}}{(R + \frac{1}{2}) (R - \frac{1}{2})}

I t s e e m s

\frac{\boldsymbol a}{\boldsymbol R} = \frac{(\boldsymbol R + \frac{1}{2})}{(\boldsymbol R - \frac{1}{2})}

e x a m p l e R = 2 \Rightarrow a = \frac{10}{3}

\dots \dots \dots .

Commented by ajfour last updated on 05/Jul/20

Answered by mr W last updated on 05/Jul/20

C e n t e r o f c i r c l e w i t h r a d i u s R i s

M (0, m)

y = x^{2}

x^{2} + {(y - m)}^{2} = R^{2}

\Rightarrow y + {(y - m)}^{2} = R^{2}

\Rightarrow y^{2} - (2 m - 1) y + m^{2} - R^{2} = 0

Δ = {(2 m - 1)}^{2} - 4 (m^{2} - R^{2}) = 0

1 + 4 R^{2} - 4 m = 0

\Rightarrow m = \frac{1}{4} + R^{2}

C e n t e r o f c i r c l e w i t h r a d i u s b i s

B (0, m + R + b)

x^{2} + {(y - m - R - b)}^{2} = b^{2}

y + {[y - (\frac{1}{4} + R^{2} + R + b)]}^{2} = b^{2}

y^{2} - 2 (R^{2} + R + b - \frac{1}{4}) y + {(R^{2} + R + b + \frac{1}{4})}^{2} - b^{2} = 0

Δ = {(R^{2} + R + b - \frac{1}{4})}^{2} - {(R^{2} + R + b + \frac{1}{4})}^{2} + b^{2} = 0

b^{2} - b - (R^{2} + R) = 0

\Rightarrow b = \frac{1}{2} (1 + 2 R + 1) = R + 1

C e n t e r o f c i r c l e w i t h r a d i u s a i s

A (2 \sqrt{R a}, m + R - a)

i t t o u c h e s t h e p a r a b o l a a t P (p, p^{2}) .

\tan θ = 2 p = \frac{2 \sqrt{R a}}{a - R}

\Rightarrow p = \frac{\sqrt{R a}}{a - R}

{(2 \sqrt{R a} - p)}^{2} + {(m + R - a - p^{2})}^{2} = a^{2}

{(2 \sqrt{R a} - \frac{\sqrt{R a}}{a - R})}^{2} + {[m + R - a - \frac{R a}{{(a - R)}^{2}}]}^{2} = a^{2}

{[2 - \frac{1}{R (\frac{a}{R} - 1)}]}^{2} (\frac{a}{R}) + \frac{1}{R^{2}} {[m - R (\frac{a}{R} - 1) - \frac{\frac{a}{R}}{{(\frac{a}{R} - 1)}^{2}}]}^{2} = {(\frac{a}{R})}^{2}

{[2 - \frac{1}{R (ξ - 1)}]}^{2} ξ + \frac{1}{R^{2}} {[\frac{1}{4} + R^{2} - R (ξ - 1) - \frac{ξ}{{(ξ - 1)}^{2}}]}^{2} = ξ^{2}

\Rightarrow ξ = \dots

C e n t e r o f c i r c l e w i t h r a d i u s c i s

C (2 \sqrt{R c}, m + R - c)

i t t o u c h e s t h e p a r a b o l a a t Q (q, q^{2}) .

\tan φ = 2 q

q = 2 \sqrt{R c} + c \sin φ = 2 \sqrt{R c} + \frac{2 c q}{\sqrt{1 + 4 q^{2}}}

\Rightarrow q - 2 \sqrt{R c} = \frac{2 c q}{\sqrt{1 + 4 q^{2}}} \dots (i)

q^{2} = m + R - c - c \cos φ = m + R - c - \frac{c}{\sqrt{1 + 4 q^{2}}}

\Rightarrow m + R - c - q^{2} = \frac{c}{\sqrt{1 + 4 q^{2}}} \dots (i i)

q - 2 \sqrt{R c} = 2 q (m + R - c - q^{2})

q^{3} - (R^{2} + R - c - \frac{1}{4}) q - \sqrt{R c} = 0

\Rightarrow q = \sqrt[3]{\frac{\sqrt{R c}}{2} + \sqrt{\frac{R c}{4} - \frac{{(R^{2} + R - c - \frac{1}{4})}^{3}}{27}}} + \sqrt[3]{\frac{\sqrt{R c}}{2} - \sqrt{\frac{R c}{4} - \frac{{(R^{2} + R - c - \frac{1}{4})}^{3}}{27}}}

p u t t h i s i n t o (i) o r (i i) t o g e t c

Commented by mr W last updated on 05/Jul/20

Commented by ajfour last updated on 05/Jul/20

b u t c i s t o b e d e t e r m i n e d, S i r .

Commented by mr W last updated on 05/Jul/20

a a n d c c a n o n l y b e d e t e r m i n e d

n u m e r i c a l l y .

Commented by ajfour last updated on 05/Jul/20

i f w e d i s c u s s, i t h i n k w e c a n g e t t h e

e x p r e s s i o n f o r r a d i i a a n d c, S i r . .

Commented by mr W last updated on 05/Jul/20

i f o u n d n o w a y t o g e t t h e m .

Commented by ajfour last updated on 06/Jul/20

S i r i g o t a = R (R + 1 / 2) / (R - 1 / 2)

E x p r e s s i o n f o r c i n d e e d s e e m s

d i f f i c u l t .

Commented by mr W last updated on 06/Jul/20

y o u g o t i t a t l e a s t f o r a! {i t}^{'} s c o r r e c t .

y o u g o t i t a t l e a s t f o r a! {i t}^{'} s c o r r e c t .

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