Circle And System Of Circles Question 176
Question: The equation of the common chord of the circles $ {{(x-a)}^{2}}+{{(y-b)}^{2}}=c^{2} $ and $ {{(x-b)}^{2}}+{{(y-a)}^{2}}=c^{2} $ is
Options:
A) $ x-y=0 $
B) $ x+y=0 $
C) $ x+y=a^{2}+b^{2} $
D) $ x-y=a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
We know that the equation of common chord is $ S_1-S_2=0 $ , where $ S_1 $ and $ S_2 $ are the equations of given circles,
therefore $ {{(x-a)}^{2}}+{{(y-b)}^{2}}+c^{2}-{{(x-b)}^{2}}-{{(y-a)}^{2}}-c^{2}=0 $
$ \Rightarrow 2bx-2ax+2ay-2by=0 $
$ \Rightarrow 2(b-a)x-2(b-a)y=0\Rightarrow x-y=0 $ .
BETA
