Circle And System Of Circles Question 327
Question: The equation of the circle having its centre on the line $ x+2y-3=0 $ and passing through the points of intersection of the circles $ x^{2}+y^{2}-2x-4y+1=0 $ and $ x^{2}+y^{2}-4x-2y+4=0 $ , is
[MNR 1992]
Options:
A) $ x^{2}+y^{2}-6x+7=0 $
B) $ x^{2}+y^{2}-3y+4=0 $
C) $ x^{2}+y^{2}-2x-2y+1=0 $
D) $ x^{2}+y^{2}+2x-4y+4=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Required circle will be $ S_1+\lambda S_2=0 $ , $ \lambda \ne -1 $ i.e., $ x^{2}+y^{2}-2x-4y+1+\lambda (x^{2}+y^{2}-4x-2y+4)=0 $
therefore $ x^{2}+y^{2}-2\frac{(1+2\lambda )}{1+\lambda }x-2\frac{(2+\lambda )}{1+\lambda }y+\frac{1+4\lambda }{1+\lambda }=0 $ Its centre $ ( \frac{1+2\lambda }{1+\lambda },\ \frac{2+\lambda }{1+\lambda } ) $ lies on $ x+2y-3=0 $ \ $ \frac{1+2\lambda }{1+\lambda }+2( \frac{2+\lambda }{1+\lambda } )-3=0 $
therefore $ \lambda =-2 $ . \ The circle is $ x^{2}+y^{2}-6x+7=0 $
BETA
