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-L-0-where-is-a-parameter-

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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