Circle And System Of Circles Question 227
Question: Let $ P(x_1,y_1) $ and $ Q(x_2,y_2) $ are two points such that their abscissa $ x_1 $ and $ x_2 $ are the roots of the equation $ x^{2}+2x-3=0 $ while the ordinates $ y_1 $ and $ y_2 $ are the roots of the equation $ y^{2}+4y-12=0 $ . The centre of the circle with PQ as diameter is
[Orissa JEE 2005]
Options:
A) $ (-1,-2) $
B) $ (1,2) $
C) $ (1,-2) $
D) $ (-1,2) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x_1,x_2 $ are roots of $ x^{2}+2x+3=0 $
therefore $ x_1+x_2=-2 $ $ \frac{x_1+x_2}{2}=-1 $
$ y_1,y_2 $ are roots of $ y^{2}+4y-12=0 $
therefore $ y_1+y_2=-4\Rightarrow \frac{y_1+y_2}{2}=-2 $ Centre of circle $ ( \frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )=(-1,-2) $ .
BETA
