circle-of-centre-P-touches-externally-both-the-circle-x-2-y-2-4x-3-0-and-x-2-y-2-6y-5-0-The-locus-of-P-is-3-4-x-2-3xy-2y-2-y-x-where-is-

Question Number 115031 by bobhans last updated on 23/Sep/20

c i r c l e o f c e n t r e P t o u c h e s e x t e r n a l l y b o t h

t h e c i r c l e x^{2} + y^{2} - 4 x + 3 = 0 a n d

x^{2} + y^{2} - 6 y + 5 = 0 . T h e l o c u s o f P i s

\frac{3}{4} x^{2} - 3 x y + 2 y^{2} = λ (y - x) w h e r e λ i s__

Answered by mr W last updated on 23/Sep/20

x^{2} + y^{2} - 4 x + 3 = 0

{(x - 2)}^{2} + y^{2} = 1^{2} \Rightarrow c e n t e r (2, 0) r a d i u s 1

x^{2} + y^{2} - 6 y + 5 = 0

x^{2} + {(y - 3)}^{2} = 2^{2} \Rightarrow c e n t e r (0, 3), r a d i u s 2

P (u, v) w i t h r a d i u s R

{(u - 2)}^{2} + v^{2} = {(R + 1)}^{2} \dots (i)

u^{2} + {(v - 3)}^{2} = {(R + 2)}^{2} \dots (i i)

(i i) - (i) :

4 (u - 1) - 3 (2 v - 3) = 2 R + 3

\Rightarrow R = 2 u - 3 v + 1

{(u - 2)}^{2} + v^{2} = {(2 u - 3 v + 2)}^{2}

u^{2} - 4 u + 4 + v^{2} = 4 u^{2} + 9 v^{2} + 4 - 12 u v + 8 u - 12 v

3 u^{2} + 8 v^{2} - 12 u v + 12 u - 12 v = 0

\frac{3}{4} u^{2} - 3 u v + 2 v^{2} = 3 (v - u)

o r

\frac{3}{4} x^{2} - 3 x y + 2 y^{2} = 3 (y - x)

\Rightarrow λ = 3

Commented by bobhans last updated on 23/Sep/20

h a h a h a \dots g a v e k u d o s s i r

Commented by bemath last updated on 23/Sep/20

Commented by bemath last updated on 23/Sep/20

w h y s i r l o c u s o f P n o t a c i r c l e ?

Commented by mr W last updated on 23/Sep/20

w h y s h o u l d i t b e a c i r c l e ?

w e c a n s e e t h e d i s t a n c e f r o m P t o

t h e g i v e n c i r c l e s c a n b e i n f i n i t e .

Commented by mr W last updated on 23/Sep/20

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

a h y p e r b o l a . j u s t h a v e a l o o k a t t h e

d e f i n i t i o n o f h y p e r b o l a .

Commented by bemath last updated on 23/Sep/20

t h e q u e s t i o n s a y :^{″} a c i r c l e o f P {c e n t r e}^{″}

Commented by mr W last updated on 23/Sep/20

P i s t h e c e n t e r o f a c i r c l e w h i c h

t a n g e n t s t w o g i v e n c i r c l e s . b u t t h e

l o c u s o f P i s n o t a c i r c l e! s a y t h e t w o

g i v e n c i r c l e s a r e

c i r c l e 1 w i t h c e n t e r A a n d r a d i u s r_{A}

c i r c l e 2 w i t h c e n t e r B a n d r a d i u s r_{B}

t h e c i r c l e w i t h c e n t e r P h a s r a d i u s R .

w e h a v e

P A = r_{A} + R

P B = r_{B} + R

P A - P B = r_{A} - r_{B} = c o n s t a n t

t h e l o c u s o f a p o i n t P, w h o s e

d i f f e r e n c e d i s t a n c e t o t w o g i v e n

p o i n t s A a n d B i s c o n s t a n t, i s a

h y p e r b o l a .

Commented by mr W last updated on 23/Sep/20

Commented by bemath last updated on 23/Sep/20

o o o i u n d e r s t o o d s i r . g a v e k u d o s

s i r

Commented by otchereabdullai@gmail.com last updated on 24/Sep/20

more blessing prof

Commented by mr W last updated on 24/Sep/20

t h a n k s!

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