from the circle, we can see that its vertex, namely (h,k), is at (0,0), h = 0, k = 0, we can also see that the radius of it is 2, r = 2, we can then just plug those values in the circle equation
[tex]\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{0}{ h},\stackrel{0}{ k})\qquad \qquad radius=\stackrel{2}{ r} \\\\[-0.35em] ~\dotfill\\\\ (x-0)^2+(y-0)^2 = 2^2\implies \sqrt{(x-0)^2+(y-0)^2} = 2[/tex]
now, for point A, we know that x = 1, so the coordinates for A(1 , y), and then if we plug that in the equation, we get
[tex](1-0)^2+(y-0)^2 = 2^2\implies \sqrt{(1-0)^2+(y-0)^2} = 2 \\\\\\ \sqrt{1^2+y^2}=2\implies \sqrt{1+y^2}=2\implies 1+y^2=2^2 \\\\\\ y^2=2^2-1\implies y^2 = 3\implies y = \sqrt{3}[/tex]
now, if we were to use the last one there, we'd get
[tex]\sqrt{(0-1)^2+(0-y)^2}=2\implies \sqrt{1+y^2}=2\implies 1+y^2=2^2 \\\\\\ y^2=2^2-1\implies y = \sqrt{3}[/tex]
