Circle And System Of Circles Question 246
Question: If polar of a circle $ x^{2}+y^{2}=a^{2} $ with respect to $ (x’,y’) $ is $ Ax+By+C=0 $ , then its pole will be
[RPET 1995]
Options:
A) $ ( \frac{a^{2}A}{-C},\frac{a^{2}B}{-C} ) $
B) $ ( \frac{a^{2}A}{C},\frac{a^{2}B}{C} ) $
C) $ ( \frac{a^{2}C}{A},\frac{a^{2}C}{B} ) $
D) $ ( \frac{a^{2}C}{-A},\frac{a^{2}C}{-B} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
Polar of the circle is $ xx’+yy’=a^{2} $ , but it is given by $ Ax+By+C=0 $ , then $ \frac{x’}{A}=\frac{y’}{B}=\frac{a^{2}}{-C} $
Hence pole is $ ( \frac{a^{2}A}{-C},\ \frac{a^{2}B}{-C} ) $ .
BETA
