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The equations of the normals to the circle x2 + y2 -8x -2y + 12 = 0 at the points whose orchnate is -1 will be (1)2x-y-7=0,2x+y-9=0 (2)2x+y-7=0,2x+y+9=0 (3)2x+y+7=0,2x+y+9=0

The equations of the normals to the circle x2 + y2 -8x -2y + 12 = 0 at the points whose orchnate is -1 will be (1)2x-y-7=0,2x+y-9=0 (2)2x+y-7=0,2x+y+9=0 (3)2x+y+7=0,2x+y+9=0

Grade:11

1 Answers

Aamina
64 Points
7 years ago
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.

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