Find $k$ if the length of the tangent segment from $(8, - 3)$ to the circle $x^{2} + y^{2} - 2 x + k y - 23 = 0$ is $\sqrt{10}$ units

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Solution

Length of the tangent segment from $(8, - 3)$ to the circle
$x^{2} + y^{2} - 2 x + k y - 23 = 0$ is
$= \sqrt{(8)^{2} + (- 3)^{2} - 2 (8) + k (- 3) - 23}$
$= \sqrt{64 + 9 - 16 - 3 k - 23}$
$= \sqrt{- 3 k + 34}$
According to the question, we have
$\sqrt{- 3 k + 34} = \sqrt{10}$
$\Rightarrow - 3 k + 34 = 10$ ....(squaring both sides)
$\Rightarrow 3 k = 24$
$\Rightarrow k = 8$

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