Circle And System Of Circles Question 187
Question: The equation of a circle passing through origin and co-axial to circles $ x^{2}+y^{2}=a^{2} $ and $ x^{2}+y^{2}+2ax=2a^{2}, $ is
Options:
A) $ x^{2}+y^{2}=1 $
B) $ x^{2}+y^{2}+2ax=0 $
C) $ x^{2}+y^{2}-2ax=0 $
D) $ x^{2}+y^{2}=2a^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of the circle which passes through origin is $ x^{2}+y^{2}+2gx+2fy=0 $ .
Radical axis with both circles is $ 2gx+2fy+a^{2}=0 $ ………….(i) $ 2(g-a)x+2fy+2a^{2}=0 $ ………….(ii)
Also radical axis of the two circles is $ x=\frac{a}{2}\Rightarrow f=0 $
From (i) and (ii), we get $ \frac{2g}{2(g-a)}=\frac{1}{2}\Rightarrow g=-a $
Hence circle is $ x^{2}+y^{2}-2ax=0 $ .
BETA
