Answer:
[tex]x=-\frac{1}{2}[/tex]
Step-by-step explanation:
equation of a circle: [tex](x - h)^2+(y-k)^2=r^2[/tex]
where (h, k) = center and r = radius
From inspection, we can see that the center of the circle = (0, 0)
and the radius = 1
[tex]\implies (x - 0)^2+(y-0)^2=1^2[/tex]
[tex]\implies x^2+y^2=1[/tex]
Substitute [tex]y=\frac{\sqrt{3}}{2}[/tex] into [tex]x^2+y^2=1[/tex] and solve for [tex]x[/tex]:
[tex]\implies x^2+(\frac{\sqrt{3}}{2})^2=1[/tex]
[tex]\implies x^2+\frac{3}{4}=1[/tex]
[tex]\implies x^2=\frac{1}{4}[/tex]
[tex]\implies x=\pm \sqrt{\frac{1}{4}}[/tex]
[tex]\implies x=\pm\frac{1}{2}[/tex]
As point P is in quadrant 2, [tex]x[/tex] is negative, so [tex]x=-\frac{1}{2}[/tex]
