Circle And System Of Circles Question 35
Question: The point of contact of the given circles $ x^{2}+y^{2}-6x-6y+10=0 $ and $ x^{2}+y^{2}=2 $ , is
Options:
A) (0, 0)
B) (1, 1)
C) (1, -1)
D) (-1, -1)
Show Answer
Answer:
Correct Answer: B
Solution:
$ x^{2}+y^{2}-6x-6y+10=0 $ …………. (i) $ x^{2}+y^{2}=2 $ …………. (ii)
$ \Rightarrow -6x-6y+12=0 $ or $ x+y-2=0 $ …………. (iii)
$ \Rightarrow x^{2}+y^{2}+2xy=4 $ {from (iii)} or $ 2xy=2 $ {from (ii)} and $ x-y=\sqrt{{{(x+y)}^{2}}-4xy}=\sqrt{4-4}=0 $ or $ x=y $ and $ x+y=2 $
$ \Rightarrow x=1,\ y=1 $ . Trick: Required point must satisfy both the circles. Obviously (1, 1) satisfies both the equations simultaneously.
BETA
