Given this unit circle, what is the value of x? (x, -7/10)(1,0)

Answer:
x = - [tex]\frac{\sqrt{51} }{10}[/tex]
Step-by-step explanation:
the equation of a circle centred at the origin is
x² + y² = r² ( r is the radius )
The radius of a unit circle is r = 1
substitute (x, - [tex]\frac{7}{10}[/tex] ) into the equation and solve for x
x² + (- [tex]\frac{7}{10}[/tex] )² = 1²
x² + [tex]\frac{49}{100}[/tex] = 1 ( subtract [tex]\frac{49}{100}[/tex] from both sides )
x² = 1 - [tex]\frac{49}{100}[/tex] = [tex]\frac{51}{100}[/tex] ( take square root of both sides )
x = ± [tex]\sqrt{\frac{51}{100} }[/tex] = ± [tex]\frac{\sqrt{51} }{10}[/tex]
since the point is in the 3rd quadrant then x < 0
x = - [tex]\frac{\sqrt{51} }{10}[/tex]