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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`
2 years ago

Aamina
64 Points
```							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.
```
2 years ago
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• Test paper with Video Solution
• Mind Map
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• NCERT Solutions
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