Circle And System Of Circles Question 266
Question: The equation of the circle which passes through the intersection of $ x^{2}+y^{2}+13x-3y=0 $ and $ 2x^{2}+2y^{2}+4x-7y-25=0 $ and whose centre lies on $ 13x+30y=0 $ is
[DCE 2001]
Options:
A) $ x^{2}+y^{2}+30x-13y-25=0 $
B) $ 4x^{2}+4y^{2}+30x-13y-25=0 $
C) $ 2x^{2}+2y^{2}+30x-13y-25=0 $
D) $ x^{2}+y^{2}+30x-13y+25=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
The equation of required circle is $ S_1+\lambda S_2=0 $ .
therefore $ x^{2}(1+\lambda )+y^{2}(1+\lambda )+x(2+13\lambda )-y( \frac{7}{2}+3\lambda )-\frac{25}{2}=0 $ Centre = $ ( \frac{-(2+13\lambda )}{2},\frac{\frac{7}{2}+3\lambda }{2} ) $
$ \because $ Centre lies on $ 13x+30y=0 $
$ \Rightarrow $ $ -13( \frac{2+13\lambda }{2} )+30( \frac{\frac{7}{2}+3\lambda }{2} )=0 $
$ \Rightarrow , $ $ \lambda =1 $ .
Hence the equation of required circle is $ 4x^{2}+4y^{2}+30x-13y-25=0. $
BETA
