abhas Active member
 Subject: 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


p123.iitr
 Subject: Re: equation of a circle solved question Tue Apr 21, 2009 2:33 pm  
 yar ek aadh question mei kya hoga ? 

abhas Active member
 Subject: Re: equation of a circle solved question Tue Apr 21, 2009 6:07 pm  
 

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 Subject: Re: equation of a circle solved question  


