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equation of a circle solved question

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abhas
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PostSubject: equation of a circle solved question Mon Nov 24, 2008 10:01 pm


Find the equation of a circle having the lines x2 + 2xy + 3x + 6y = 0 as its normals and having size just
sufficient to contain the circle x (x 4) + y (y 3) = 0

Solution
On factorising the equation of the pair of straight lines x2 + 2xy + 3x + 6y = 0, we get :
(x + 2y) (x + 3) = 0
⇒ Two normals are x = 2y ...........(i) and x = 3 .........(ii)
The point of intersection of normals (i) and (ii) is centre of the required circle as centre lies on all normal
lines.
Solving (i) and (ii), we get :
centre ≡ C1 ≡ (3, 3/2)
Given circle is C2 ≡` x (x 4) + y(y 3) = 0 ⇒ x2 + y2 4x 3y = 0
⇒ centre ≡ C2 ≡ (2, 3/2) and radius = r = 5/2
If the required circle just contains the given circle, the given circle should touch the required circle internally
from inside.
⇒ radius of the required circle = |C1 C2| + r
⇒ radius of the required circle = 5 + 5/2 = 15/2
Hence, equation of required circle is (x + 3)2 + (y 3/2)2 = 225/4

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PostSubject: Re: equation of a circle solved question Tue Apr 21, 2009 2:33 pm

yar ek aadh question mei kya hoga ?
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abhas
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PostSubject: Re: equation of a circle solved question Tue Apr 21, 2009 6:07 pm

lol!
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