Circle And System Of Circles Question 18
Question: If the circle $ x^{2}+y^{2}=4 $ bisects the circumference of the circle $ x^{2}+y^{2}-2x+6y+a=0 $ , then a equals
[RPET 1999]
Options:
A) 4
B) -4
C) 16
D) -16
Show Answer
Answer:
Correct Answer: C
Solution:
The common chord of given circles is $ 2x-6y-4-a=0 $ …..(i) Since, $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,
therefore, (i) passes through the centre of second circle i.e., (1, - 3). \ 2 + 18 - 4 -a = 0
therefore a = 16.
BETA
