Circle And System Of Circles Question 287
Question: The points of intersection of the circles $ x^{2}+y^{2}=25 $ and $ x^{2}+y^{2}-8x+7=0 $ are
[MP PET 1988]
Options:
A) (4, 3) and (4, -3)
B) (4, -3) and (-4, -3)
C) (-4, 3) and (4, 3)
D) (4, 3) and (3, 4)
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{2}+y^{2}=25 $ ………….. (i) $ x^{2}+y^{2}-8x+7=0 $ ………….. (ii)
From (i) and (ii), we get $ 8x=7+25 $ or $ x=4 $ and for $ x=4 $ , we get $ y=\pm 3 $ .
Hence points of intersection are (4, 3), (4, -3).
BETA
