The number of normal line to the circle $x^{2} + y^{2} - 14 x - 2 y + 5 = 0$ at a point whose abscissa is $10$ is/are

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Solution

The correct option is D $2$
Putting $x = 10$ in the given circle,
$\Rightarrow 100 + y^{2} - 140 - 2 y + 5 = 0 \Rightarrow y^{2} - 2 y - 35 = 0$
$\Rightarrow (y - 7) (y + 5) = 0 \Rightarrow y = 7, - 5$
Hence the points are $(10, - 5)$ and $(10, 7)$
Also these points are not end point of any diameter to the circle
Hence there will be two normal from these two points can be drawn.

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