For what value of $K$ is the point $(K, - 2)$ outside the circle with equation $x^{2} + y^{2} = 9$ ?

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Solution

Since the point $(k, - 2)$ is outside the circle with equation $x^{2} + y^{2} = 9$ ,

Therefore, substitute $x = K$ and $y = - 2$ in the equation $x^{2} + y^{2} = 9$ as shown below:

$(K)^{2} + (- 2)^{2} = 9$
$K^{2} + 4 = 9$
$K^{2} = 9 - 4$
$K^{2} = 5$
$K = \pm \sqrt{5}$

Hence, $(- i n f i n i t y, - \sqrt{5}) \cup (\sqrt{5}, + i n f i n i t y)$

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