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 = λ(y−x) where λ is _


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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

	$x^{2} + y^{2} - 4 x + 3 = 0$ ${(x - 2)}^{2} + y^{2} = 1^{2} \Rightarrow c e n t e r (2, 0) r a d i u s 1$ $x^{2} + y^{2} - 6 y + 5 = 0$ $x^{2} + {(y - 3)}^{2} = 2^{2} \Rightarrow c e n t e r (0, 3), r a d i u s 2$ $P (u, v) w i t h r a d i u s R$ ${(u - 2)}^{2} + v^{2} = {(R + 1)}^{2} . . . (i)$ $u^{2} + {(v - 3)}^{2} = {(R + 2)}^{2} . . . (i i)$ $(i i) - (i) :$ $4 (u - 1) - 3 (2 v - 3) = 2 R + 3$ $\Rightarrow R = 2 u - 3 v + 1$ ${(u - 2)}^{2} + v^{2} = {(2 u - 3 v + 2)}^{2}$ $u^{2} - 4 u + 4 + v^{2} = 4 u^{2} + 9 v^{2} + 4 - 12 u v + 8 u - 12 v$ $3 u^{2} + 8 v^{2} - 12 u v + 12 u - 12 v = 0$ $\frac{3}{4} u^{2} - 3 u v + 2 v^{2} = 3 (v - u)$ $o r$ $\frac{3}{4} x^{2} - 3 x y + 2 y^{2} = 3 (y - x)$ $\Rightarrow λ = 3$
	Commented by bobhans last updated on 23/Sep/20

	$h a h a h a . . . g a v e k u d o s s i r$
	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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