Question

Find the points of intersection of the graph of the following equations:
y = (sqrt3/3x) + 1
x^2 + (y-1)^2 = 4

Answer 1

\[y=\frac{\sqrt 3}{3x}+1\]?

Answer 2

That is,
\[y = \frac{ \sqrt3 }{ 3 }x + 1 \]
and
\[x^2 + (y-1)^2 = 4\]

Answer 3

Hints:
Given:
\(y = \frac{ \sqrt3 }{ 3 }x + 1\)
substitute into the second equation:
\(x^2 + (y-1)^2 = 4\)
\(x^2 + ( \frac{ \sqrt3 }{ 3 }x + 1-1)^2 = 4\)
Simplify and solve for x (two roots, symm. about y-axis)