Circle And System Of Circles Question 328
Question: The point at which the normal to the circle $ x^{2}+y^{2}+4x+6y-39=0 $ at the point (2, 3) will meet the circle again, is
Options:
A) (6, -9)
B) (6, 9)
C) (-6, -9)
D) (-6, 9)
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of normal will be $ \frac{x-2}{2+2}=\frac{y-3}{3+3} $
$ \Rightarrow $ $ 3x-2y=0\Rightarrow x=\frac{2y}{3} $ Thus $ {{( \frac{2y}{3} )}^{2}}+y^{2}+4( \frac{2y}{3} )+6y-39=0 $
$ \Rightarrow y=3,\ -9\Rightarrow x=2,\ -6 $
Hence another point will be $ (-6,,-9) $ .
BETA
