Circle And System Of Circles Question 297
Question: If the centre of a circle which passing through the points of intersection of the circles $ x^{2}+y^{2}-6x+2y+4=0 $ and $ x^{2}+y^{2}+2x-4y-6=0 $ is on the line $ y=x $ , then the equation of the circle is
[RPET 1991; Roorkee 1989]
Options:
A) $ 7x^{2}+7y^{2}-10x+10y-11=0 $
B) $ 7x^{2}+7y^{2}+10x-10y-12=0 $
C) $ 7x^{2}+7y^{2}-10x-10y-12=0 $
D) $ 7x^{2}+7y^{2}-10x-12=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Family of circles through points of intersection of two circles is $ S_1+\lambda S_2,(\lambda \ne -1) $ . $ x^{2}+y^{2}-6x+2y+4+\lambda (x^{2}+y^{2}+2x-4y-6)=0 $
Centre is $ (3-\lambda ,\ -1+2\lambda ) $ .
It lies on $ y=x $ .
Therefore, $ -1+2\lambda =3-\lambda \Rightarrow \lambda =\frac{4}{3} $
Hence equation of circle can be found by substituting $ \lambda $ in the family of circles above.
BETA
