The number of integral values of $k$ for which the chord of the circle $x^{2} + y^{2} = 125$ passing through $P (8, k)$ gets bisected at $P (8, k)$ and has integral slope, is

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Solution

Given circle is $x^{2} + y^{2} = 125$ and mid point of chord is $P (8, k)$
Equation of chord with mid point $P (8, k)$ is $T = S_{1}$
$\Rightarrow 8 x + k y - 64 - k^{2} = 0$
Slope of chord is, $m = - \frac{8}{k}$
As $m$ is an integer,
so $k = \pm 1, \pm 2, \pm 4, \pm 8$
But $(8, k)$ must also lie inside the circle $x^{2} + y^{2} = 125.$
$\Rightarrow 64 + k^{2} - 125 < 0$
$\Rightarrow k^{2} < 61$
$k$ can be $\pm 1, \pm 2, \pm 4$
$\Rightarrow 6$ values

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The number of integral values of k for which the chord of the circle x2+y2=125 passing through P(8,k) gets bisected at P(8,k) and has integral slope, is

The number of integral values of $k$ for which the chord of the circle $x^{2} + y^{2} = 125$ passing through $P (8, k)$ gets bisected at $P (8, k)$ and has integral slope, is