The normal to the circle x2+y2+2x−10y+k = 0 which is perpendicular to x-3y+2=0 is
x+3y+2=0
3x+y=0
3x+y=2
3x+y=4
x2+y2+2x−10y+k = 0 ⊥ to x−3y+2 = 0
C = (-1,5)
Y−5 = -3(X - 1) m = 13 perpendicular slope = -3
3x + y -2 = 0
The normal to the circle x2+y2+2x−10y+k = 0 which is perpendicular to x-3y+2=0 is