Circle And System Of Circles Question 215
Question: If the lines $ 2x+3y+1=0 $ and $ 3x-y-4=0 $ lie along diameters of a circle of circumference $ 10\pi $ , then the equation of the circle is
[AIEEE 2004]
Options:
A) $ x^{2}+y^{2}+2x-2y-23=0 $
B) $ x^{2}+y^{2}-2x-2y-23=0 $
C) $ x^{2}+y^{2}+2x-2y-23=0 $
D) $ x^{2}+y^{2}-2x+2y-23=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
According to question two diameters of the circle are $ 2x+3y+1=0 $ and $ 3x-y-4=0 $
Solving, we get $ x=1,,y=-1 $
$ \therefore $ Centre of the circle is (1, - 1) Given $ 2\pi r=10\pi \Rightarrow r=5 $
$ \therefore $ Required circle is $ {{(x-1)}^{2}}+{{(y+1)}^{2}}=5^{2} $ or $ x^{2}+y^{2}-2x+2y-23=0 $ .
BETA
