Circle And System Of Circles Question 110
Question: Centre of circle $ (x-x_1)(x-x_2) $ $ +(y-y_1)(y-y_2) $ $ =0 $ is
Options:
A) $ ( \frac{x_1+y_1}{2},\ \frac{x_2+y_2}{2} ) $
B) $ ( \frac{x_1-y_1}{2},\ \frac{x_2-y_2}{2} ) $
C) $ ( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2} ) $
D) $ ( \frac{x_1-x_2}{2},\ \frac{y_1-y_2}{2} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ (x_1,\ y_1) $ and $ (x_2,\ y_2) $ are extreme points of diameter.
Hence centre is $ ( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2} ) $ .
BETA
