Circle And System Of Circles Question 29
Question: The equation of a circle passing through points of intersection of the circles $ x^{2}+y^{2}+13x-3y=0 $ and $ 2x^{2}+2y^{2}+4x-7y-25=0 $ and point (1, 1) is
[RPET 1988, 89; IIT 1983]
Options:
A) $ 4x^{2}+4y^{2}-30x-10y-25=0 $
B) $ 4x^{2}+4y^{2}+30x-13y-25=0 $
C) $ 4x^{2}+4y^{2}-17x-10y+25=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Required equation is $ (x^{2}+y^{2}+13x-3y)+\lambda (2x^{2}+2y^{2}+4x-7y-25)=0 $ which passes through (1, 1), so $ \lambda =\frac{1}{2} $ .
Hence required equation is $ 4x^{2}+4y^{2}+30x-13y-25=0 $ .
BETA
