Question

For what value of $K$ is the point $(1,K)$ inside the circle with equation $x^2+y^2=4$?

Answer 1

Since the point $(1,K)$ is inside the circle with equation $x^2+y^2=4$,

Therefore, substitute $x=1$ and $y=K$ in the equation $x^2+y^2=4$ as shown below:

$(1{)}^2+(K{)}^2=4$

$\Rightarrow 1+K^2=4$

$\Rightarrow K^2=4-1$

$\Rightarrow K^2=3$

$\Rightarrow K=\pm \sqrt{3}$

Hence, $-\sqrt{3}<K<\sqrt{3}$