prove that equation of a circle passing through the points of intersection of a circle S=0 and a line L=0 can be taken as S+λL=0 where λ is a parameter


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Question Number 147144 by gsk2684 last updated on 18/Jul/21

$p r o v e t h a t$ $e q u a t i o n o f a c i r c l e p a s s i n g t h r o u g h$ $t h e p o i n t s o f i n t e r s e c t i o n o f a c i r c l e$ $S = 0 a n d a l i n e L = 0 c a n b e t a k e n a s$ $S + λ L = 0 w h e r e λ i s a p a r a m e t e r$
	Answered by mr W last updated on 18/Jul/21

	$s a y i n t e r s e c t i o n a t (a, b)$ $S (a, b) = 0$ $L (a, b) = 0$ $S (a, b) + λ L (a, b) = 0$ $\Rightarrow S (x, y) + λ L (x, y) = 0 i s a c i r c l e$ $p a s s i n g t h r o u g h (a, b) .$
	Commented by gsk2684 last updated on 20/Jul/21

	$o f c o u r s e w h e n t w o c u r v e s i n t e r s e c t$ $a t a p o i n t P t h e n i t s a t i f i e s b o t h$ $t h e e q u a t i o n .$ $i n e e d t o k n o w h o w c a n w e p r o v e$ $t h a t g e n e r a l f o r m r e p r e s e n t a c i r c l e ?$
	Commented by mr W last updated on 20/Jul/21

	$i f S (x, y) = 0 i s a c i r c l e, t h e n$ $S (x, y) = {A x}^{2} + B x + {A y}^{2} + C y + D = 0$ $i s L (x, y) = 0 i s a l i n e, t h e n$ $L (x, y) = K x + H y + G = 0$ $S (x, y) + λ L (x, y) = 0$ ${A x}^{2} + (B + λ K) x + {A y}^{2} + (C + λ H) y + D + λ G = 0$ $t h i s i s a l s o a c i r c l e .$
	Commented by gsk2684 last updated on 21/Jul/21

	$c o n d i t i o n t h a t t h e e q u a t i o n$ ${a x}^{2} + {a y}^{2} + 2 g x + 2 f y + c = 0 r e p r e s e n t$ $a c i r c l e i s g^{2} + f^{2} - a c ⩾ 0 ∵ r a d i u s ⩾ 0$ $i w o u l d l i k e t o v e r i f y t h e c o n d i t i o n$ ${(\frac{B + λ K}{2})}^{2} + {(\frac{C + λ H}{2})}^{2} - A (D + λ G) ⩾ 0$ $c o n s i d e r$ ${(\frac{B + λ K}{2})}^{2} + {(\frac{C + λ H}{2})}^{2} - A (D + λ G)$ $[{(\frac{B}{2})}^{2} + {(\frac{C}{2})}^{2} - A D] + λ (B K + C H - A G) + \frac{λ^{2} (K^{2} + H^{2})}{4}$ $s o m e c o n f u s i o n t o p r o c e e d f r o m t h i s$
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