If $x^{2} + y^{2} - 2 k x - 2 k y + 3 k^{2} - 6 k + 8 = 0$ is a real circle, then the possible value(s) of $k$ is/are

2

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3

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4

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5

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Solution

The correct options are
A $2$
B $3$
C $4$
Given circle equation
$x^{2} + y^{2} - 2 k x - 2 k y + 3 k^{2} - 6 k + 8 = 0$
$\Rightarrow g = - k, f = - k, c = 3 k^{2} - 6 k + 8$

For real circle,
Radius of the circle $r \geq 0$
$\Rightarrow g^{2} + f^{2} - c \geq 0 \Rightarrow k^{2} + k^{2} - (3 k^{2} - 6 k + 8) \geq 0 \Rightarrow - k^{2} + 6 k - 8 \geq 0 \Rightarrow k^{2} - 6 k + 8 \leq 0 \Rightarrow (k - 2) (k - 4) \leq 0 ∴ 2 \leq k \leq 4$

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