The normal to the circle $x^{2} + y^{2} + 2 x - 10 y + k$ = 0 which is perpendicular to x-3y+2=0 is

x+3y+2=0

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3x+y=0

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3x+y=2

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3x+y=4

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Solution

The correct option is C
3x+y=2

$x^{2} + y^{2} + 2 x - 10 y + k$ = 0 $⊥$ to $x - 3 y + 2$ = 0

C = (-1,5)

$Y - 5$ = -3(X - 1) m = $\frac{1}{3}$ perpendicular slope = -3

3x + y -2 = 0

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