Circle And System Of Circles Question 190
Question: The equation of the normal to the circle $ x^{2}+y^{2}=9 $ at the point $ ( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) $ is
Options:
A) $ x+y=0 $
B) $ x-y=\frac{\sqrt{2}}{3} $
C) $ x-y=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We know that the equation of normal to the circle $ x^{2}+y^{2}=a^{2} $ at the point $ (x_1,\ y_1) $ is $ \frac{x}{x_1}-\frac{y}{y_1}=0 $ .
Therefore $ \frac{x}{1/\sqrt{2}}-\frac{y}{1/\sqrt{2}}=0\Rightarrow x-y=0 $ .
BETA
