The given equation of the circle is x2+y2-8x-2y+12=0.
The points on the circle with ordinate -1 can be obtained by substituting y=-1 in the circle equation.
By substituting y=-1 we get aquadratic equation in x.
i.e, x2-8x+15=0
x2-5x-3x+15=0
x=3 and x=5
The points on the circle with ordinate -1 are (3,-1) and (5,-1).
All the normals to the circle pass through the centre of the circle
The centre of the given circle is (4,1)
Equation of normal at (3,-1) passes through (3,-1) and (4,1)
From the formula
y-y1=[y2-y1 /x2-x1](x-x1)
The equation comes out to be as 2x-y-7=0
Similarly from the points (4,1) and (5,-1)
The equation of another normal comes out to be 2x+y-9=0
Answer is the first option.
HOPE THIS HELPS.
