Answer:
[tex] x^2 + \frac{1}{49} =1 [/tex]
And solving for x we got:
[tex] x^2 = 1- \frac{1}{49}[/tex]
[tex] x^2 = \frac{48}{49}[/tex]
And taking square root we got:
[tex] x = \sqrt{\frac{48}{49}}= \frac{\sqrt{48}}{7} = \frac{4\sqrt{3}}{7}[/tex]
Step-by-step explanation:
For this case we have the following point given [tex] P(x , -1/7)[/tex]
And we want to find the value of x, since P lies on the unitray circle if we find the distance from P to the center of the unitary circle (0,0) we need to get 1. Using the definition of Euclidean distance that means:
[tex] d = \sqrt{(x -0)^2 +(-1/7 -0)^2}=1[/tex]
And if we square both sides of the last equation we got:
[tex] x^2 + \frac{1}{49} =1 [/tex]
And solving for x we got:
[tex] x^2 = 1- \frac{1}{49}[/tex]
[tex] x^2 = \frac{48}{49}[/tex]
And taking square root we got:
[tex] x = \sqrt{\frac{48}{49}}= \frac{\sqrt{48}}{7} = \frac{4\sqrt{3}}{7}[/tex]