Circle And System Of Circles Question 321
Question: The equations of the normals to the circle $ x^{2}+y^{2}-8x-2y+12=0 $ at the points whose ordinate is -1, will be
Options:
A) $ 2x-y-7=0,,2x+y-9=0 $
B) $ 2x+y+7=0,,2x+y+9=0 $
C) $ 2x+y-7=0,2x+y+9=0 $
D) $ 2x-y+7=0,,2x-y+9=0 $
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Answer:
Correct Answer: A
Solution:
The abscissa of point is found by substituting the ordinates and
Solving for abscissa.
$ \Rightarrow x^{2}-8x+15=0 $
$ \Rightarrow x=\frac{8\pm \sqrt{64-60}}{2}=\frac{8\pm 2}{2}=5 $ or 3 i.e., points are $ (5,\ -1) $ and (3,-1). Normal is given by, $ \frac{x-5}{5-4}=\frac{y+1}{-1-1}\Rightarrow 2x+y-9=0 $ and $ \frac{x-3}{3-4}=\frac{y+1}{-1-1}\Rightarrow 2x-y-7=0 $ .
BETA
