determine equation of circle that offensive the both of coordinate and through (2,−1)


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Question Number 5998 by Kasih last updated on 09/Jun/16

$d e t e r m i n e e q u a t i o n o f c i r c l e t h a t o f f e n s i v e t h e$ $b o t h o f c o o r d i n a t e a n d t h r o u g h (2, - 1)$
Commented by Rasheed Soomro last updated on 09/Jun/16

$W h a t d o y o u m e a n b y ‘ ‘ {o f f e n s i v e}^{″} ?$
Commented by Kasih last updated on 09/Jun/16

$i {d o n}^{'} t k n o w . m a y b e o f f e n d, t r e a t, t o u c h, o r p a s s .$ $s o r r y i {c a n}^{'} t u s e e n g l i s h l a n g u a g e w e l l$
Commented by FilupSmith last updated on 09/Jun/16

$Assumed Correction :$ $Determin the equation of the circle$ $that passes through the coordinate$ $(2, - 1)$
Commented by Kasih last updated on 09/Jun/16

$a n d p a s s t h r o u g h t h e b o t h a x i s i n c a r t r s i a n c o o r d i n a t e$
Commented by prakash jain last updated on 09/Jun/16

$Mostly the question might be 2 axis as tangent$ $and passing thru the given point .$
	Answered by Rasheed Soomro last updated on 10/Jun/16

	$∙ T h e c e n t e r o f t h e c i r c l e w h i c h t o u c h e s$ $b o t h a x e s i s e q u i d i s t a n t f r o m t h e a x e s .$ $∙ A l s o i f t h e c i r c l e i s t a n g e n t t o b o t h a x e s,$ $i t i s l i m e t e d t o o n e q u a r d a n t o n l y .$ $∙ S i n c e i t p a s s e s t h r o u g h (2, - 1) t h a t m e a n s t h e$ $c i r c l e i s i n f o u r t h q u a r d a n t .$ $∙ L e t h i s t h e d i r e c t e d d i s t a n c e o f t h e c e n t r e f r o m$ $x - a x i s . T h e n t h e d i r e c t e d d i s t a n c e o f t h e c e n t e r$ $f r o m y - a x i s i s - h . [y - c o o r d i n a t e i s n e g a t i v e i n$ $f o u r t h q u a r d a n t]$ $∙ T h e r a d i u s o f t h e c i r c l e i s ∣ h ∣ .$ $H e n c e t h e e q u a t i o n o f s u c h c i r c l e w i l l b e :$ ${(x - h)}^{2} + {(y - (- h))}^{2} =∣ h ∣^{2}$ $A s t h i s c i r c l e p a s s e s t h r o u g h (2, - 1), t h e$ $e q u a t i o n w i l l b e s a t i s f i e d b y (x, y) = (2, - 1) :$ ${(2 - h)}^{2} + {(- 1 + h)}^{2} = h^{2} ∵ ∣ h ∣^{2} = h^{2}$ $4 - 4 h + h^{2} + 1 - 2 h + h^{2} - h^{2} = 0$ $h^{2} - 6 h + 5 = 0$ $(h - 1) (h - 5) = 0$ $h = 1 \lor h = 5$ $T h e e q u a t i o n o f t h e r e q u i r e d c i r c l e i s$ ${(x - 1)}^{2} + {(y + 1)}^{2} = 1$ $O r {(x - 5)}^{2} + {(y + 5)}^{2} = 25$
	Commented by Rasheed Soomro last updated on 10/Jun/16

	Commented by Rasheed Soomro last updated on 10/Jun/16

	Commented by Kasih last updated on 10/Jun/16

	$t h a n k y o u$
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