determine-equation-of-circle-that-offensive-the-both-of-coordinate-and-through-2-1-

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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