Circle And System Of Circles Question 291
Question: The points of intersection of circles $ x^{2}+y^{2}=2ax $ and $ x^{2}+y^{2}=2by $ are
[AMU 2000]
Options:
A) (0, 0), (a, b)
B) (0, 0), $ ( \frac{2ab^{2}}{a^{2}+b^{2}},\frac{2ba^{2}}{a^{2}+b^{2}} ) $
C) (0, 0), $ ( \frac{a^{2}+b^{2}}{a^{2}},\frac{a^{2}+b^{2}}{b^{2}} ) $
D) None of the above
Show Answer
Answer:
Correct Answer: B
Solution:
Given circles are $ x^{2}+y^{2}=2ax $ …………..(i) and $ x^{2}+y^{2}=2by $ …………..(ii) (i) - (ii)
therefore $ 0=2(ax-by) $
therefore $ y=\frac{a}{b}x $
From (i), $ x^{2}+\frac{a^{2}}{b^{2}}x^{2}=2ax $
therefore $ x{ ( 1+\frac{a^{2}}{b^{2}} )x-2a }=0 $
therefore $ x=0,\ \frac{2ab^{2}}{a^{2}+b^{2}} $
For $ x=0 $ , $ y=0 $ and for $ x=\frac{2ab^{2}}{a^{2}+b^{2}} $ , $ y=\frac{2a^{2}b}{a^{2}+b^{2}} $ \
The points of intersection are (0, 0) and $ ( \frac{2ab^{2}}{a^{2}+b^{2}},\ \frac{2a^{2}b}{a^{2}+b^{2}} ) $ .
BETA
