Prove that the locus of a point which moves its distance from the point (−b, 0) is p times its distance from the point (b, 0) is (p^2 − 1)(x^2 + y^2 + b^2 ) − 2b(p^2 + 1)x = 0 Show that this locus is a circle and find its radius.


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Question Number 6716 by Tawakalitu. last updated on 15/Jul/16

$P r o v e t h a t t h e l o c u s o f a p o i n t w h i c h m o v e s i t s d i s t a n c e f r o m$ $t h e p o i n t (- b, 0) i s p t i m e s i t s d i s t a n c e f r o m t h e p o i n t (b, 0) i s$ $(p^{2} - 1) (x^{2} + y^{2} + b^{2}) - 2 b (p^{2} + 1) x = 0$ $S h o w t h a t t h i s l o c u s i s a c i r c l e a n d f i n d i t s r a d i u s .$
	Answered by Yozzii last updated on 16/Jul/16

	$L e t p o i n t s P, A a n d B h a v e c o o r d i n a t e s (x, y), (- b, 0)$ $a n d (b, 0) r e s p e c t i v e l y .$ $T h e i n f o r m a t i o n g i v e s ∣ P A ∣= p ∣ P B ∣ w h e r e p > 0 .$ $∣ P A ∣= \sqrt{{(x + b)}^{2} + y^{2}} a n d ∣ P B ∣= \sqrt{{(x - b)}^{2} + y^{2}} .$ $∴ \sqrt{{(x + b)}^{2} + y^{2}} = p \sqrt{{(x - b)}^{2} + y^{2}}$ $\Rightarrow {(\sqrt{{(x + b)}^{2} + y^{2}})}^{2} = {(p \sqrt{{(x - b)}^{2} + y^{2}})}^{2}$ ${(x + b)}^{2} + y^{2} = p^{2} {(x - b)}^{2} + p^{2} y^{2}$ $x^{2} + 2 b x + b^{2} + y^{2} = p^{2} x^{2} - 2 p^{2} b x + p^{2} b^{2} + p^{2} y^{2}$ $(p^{2} - 1) x^{2} + (p^{2} - 1) y^{2} + (p^{2} - 1) b^{2} - 2 b x (1 + p^{2}) = 0$ $(p^{2} - 1) (x^{2} + y^{2} + b^{2}) - 2 b (p^{2} + 1) x = 0 . . . . (1)$ $E q u a t i o n (1) r e p r e s e n t s t h e l o c u s o f P$ $i n r e c t a n g u l a r c o o r d i n a t e s .$ $- - - - - - - - - - - - - - - - - - - - - - - -$ $I f p \neq \pm 1, w e c a n r e w r i t e (1) a s$ $x^{2} + y^{2} + b^{2} - 2 (\frac{b (p^{2} + 1)}{p^{2} - 1}) x = 0$ $x^{2} - 2 (\frac{b (p^{2} + 1)}{p^{2} - 1}) x + {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} - {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2} = 0$ ${(x - \frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2} = {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2}$ $\Rightarrow \sqrt{{(x - \frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2}} =∣ \frac{b}{p^{2} - 1} ∣ (p^{2} + 1) . . . . (2)$ $T h e f o r m o f (2) i m p l i e s t h a t t h e d i s t a n c e$ $b e t w e e n P (x, y) a n d t h e f i x e d p o i n t (\frac{b (p^{2} + 1)}{p^{2} - 1}, 0)$ $i s a l w a y s c o n s t a n t a s P m o v e s a n d b, p a r e$ $c o n s t a n t s . H e n c e, t h e l o c u s o f P i s a$ $c i r c l e a n d i t s r a d i u s i s ∣ \frac{b}{p^{2} - 1} ∣ (p^{2} + 1) .$
	Commented by Tawakalitu. last updated on 16/Jul/16

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