Circle And System Of Circles Question 355
The normal to the circle $ x^{2}+y^{2}-3x-6y-10=0 $ at the point (3, -4), is
[RPET 1986, 89]
Options:
A) $ 2x+9y-30=0 $
B) $ 9x-2y+35=0 $
C) $ 2x-9y+30=0 $
D) $ 2x-9y-30=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{x-x_1}{x_1+g}=\frac{y-y_1}{y_1+f} $
$ \Rightarrow \frac{x+3}{-3-\frac{3}{2}}=\frac{y-4}{4-3}\Rightarrow 2x+9y-30=0 $ .
BETA
