[tex]\bf \begin{array}{rllll}
p(x&,&\frac{1}{9})\\
\uparrow &&\uparrow \\ a&&b \end{array}\impliedby \textit{is on the unit circle}[/tex]
well, the Unit Circle is called like so, because its radius/hypotenuse is 1unit
so, we know c = 1
now, simply let's use the pythagorean theorem to get the "x" coordinate then
[tex]\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\implies \pm\sqrt{1^2-\left( \frac{1}{9} \right)^2}=a
\\\\\\
\pm\sqrt{1-\frac{1}{81}}=a\implies \pm\sqrt{\frac{80}{81}}=a\implies \pm\cfrac{\sqrt{80}}{\sqrt{81}}=a\implies \pm\cfrac{\sqrt{4^2\cdot 5}}{\sqrt{9^2}}=a
\\\\\\
\pm\cfrac{4\sqrt{5}}{9}=a[/tex]
but hmmm which one is it? the +/-? well, we know the terminal point is on the II quadrant, and on that quadrant, the value for "x" is negative
thus [tex]\bf -\cfrac{4\sqrt{5}}{9}=a[/tex]
