The equation of the normals to the circle $x^{2} + y^{2} - 8 x - 2 y + 12 = 0$ at the points whose ordinate is $- 1$ is/are

2 x + y - 9 = 0

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2 x - y - 7 = 0

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2 x + y - 7 = 0

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2 x - y - 9 = 0

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Solution

The correct options are
A $2 x + y - 9 = 0$
B $2 x - y - 7 = 0$
Given circle is $x^{2} + y^{2} - 8 x - 2 y + 12 = 0$
Putting the ordinate, we get the abscissa
$x^{2} - 8 x + 15 = 0 \Rightarrow (x - 5) (x - 3) = 0 \Rightarrow x = 3, 5$
$∴$ Points are $(5, - 1)$ and $(3, - 1)$

Centre of the circle is $(4, 1)$
Normal is given by
$\frac{x - 5}{5 - 4} = \frac{y + 1}{- 1 - 1} \Rightarrow 2 x + y - 9 = 0$
And
$\frac{x - 3}{3 - 4} = \frac{y + 1}{- 1 - 1} \Rightarrow 2 x - y - 7 = 0$

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