Question-159392

Question Number 159392 by ajfour last updated on 16/Nov/21

Commented by ajfour last updated on 16/Nov/21

F i n d r a d i u s o f t h e c i r c l e s (e q u a l) .

Answered by mr W last updated on 17/Nov/21

Commented by mr W last updated on 18/Nov/21

e q n . o f O C :

s a y y = m x \Rightarrow m x - y = 0

e q n . o f A D :

s a u y = - k (x - 1) \Rightarrow k x + y - k = 0

P (r, 1 - r)

Q (q, r)

r = - \frac{m r - 1 + r}{\sqrt{1 + m^{2}}} \dots (i)

r = \frac{m q - r}{\sqrt{1 + m^{2}}} \dots (i i)

r = - \frac{k r + 1 - r - k}{\sqrt{1 + k^{2}}} \dots (i i i)

r = - \frac{k q + r - k}{\sqrt{1 + k^{2}}} \dots (i v)

4 u n k n o w n s : r, m, k, q

m q - r + m r - 1 + r = 0

\Rightarrow q = \frac{1}{m} - r

k q + r - k = k r + 1 - r - k

\Rightarrow k = \frac{1 - 2 r}{q - r}

k = \frac{1 - 2 r}{\frac{1}{m} - 2 r} \dots (I)

f r o m (i i) :

r = \frac{1 - (m + 1) r}{\sqrt{1 + m^{2}}}

r = \frac{1}{1 + m + \sqrt{1 + m^{2}}} \dots (I I)

f r o m (i v) :

1 - k + \sqrt{1 + k^{2}} = (1 - \frac{1}{m}) \frac{k}{r} \dots (I I I)

p u t t i n g I, I I i n t o I I I w e g e t a n . e q n .

f o r m .

\Rightarrow m \approx 1.243756

\Rightarrow r \approx 0.2694

Commented by mr W last updated on 18/Nov/21

Commented by ajfour last updated on 18/Nov/21

E x c e l l e n t s i r, t h a n k s!

i m o u t o f m t o w n t o t a k e u p a j o b,

h e n c e m y i n c o n s i s t e n c y . .

Commented by mr W last updated on 18/Nov/21

g o o d l u c k w i t h t h e n e w j o b s i r!

Commented by Tawa11 last updated on 28/Nov/21

Great sir

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