Circle And System Of Circles Question 234
Question: The equation of the circle through the point of intersection of the circles $ x^{2}+y^{2}-8x-2y+7=0 $ , $ x^{2}+y^{2}-4x+10y+8=0 $ and (3, -3) is
Options:
A) $ 23x^{2}+23y^{2}-156x+38y+168=0 $
B) $ 23x^{2}+23y^{2}+156x+38y+168=0 $
C) $ x^{2}+y^{2}+156x+38y+168=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of circle is $ (x^{2}+y^{2}-8x-2y+7)+\lambda (x^{2}+y^{2}-4x+10y+8)=0 $
Also point $ (3,\ -3) $ lies on the above equation.
$ \Rightarrow \lambda =\frac{7}{16} $
Hence required equation is $ 23x^{2}+23y^{2}-156x+38y+168=0 $ .
BETA
