Circle And System Of Circles Question 316
Question: The polar of the point (5, -1/2) w.r.t circle $ {{(x-2)}^{2}}+y^{2}=4 $ is
[RPET 1996]
Options:
A) $ 5x-10y+2=0 $
B) $ 6x-y-20=0 $
C) $ 10x-y-10=0 $
D) $ x-10y-2=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
The polar of the point $ ( 5,\ -\frac{1}{2} ) $ is $ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 $
$ \Rightarrow 5x-\frac{1}{2}y-2(x+5)+0+0=0 $
$ \Rightarrow 3x-\frac{y}{2}-10=0\Rightarrow 6x-y-20=0 $ .
BETA
