Question

which graph represents the solution of the system \\( \left\\{ \

$$\begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 4 } \\\\ { x - y = 1 } \\end{array}$$

\
ight. \\)?

Explanation:

Step1: Analyze the equations

The equation \(x^{2}+y^{2}=4\) represents a circle with center \((0,0)\) and radius \(r = 2\) (since \(r^{2}=4\)). The equation \(x - y=1\) can be rewritten as \(y=x - 1\), which is a straight line with slope \(m = 1\) and \(y\)-intercept \(b=-1\).

Step2: Find the intersection points (optional for graph - recognition)

Substitute \(y=x - 1\) into \(x^{2}+y^{2}=4\). We get \(x^{2}+(x - 1)^{2}=4\). Expand: \(x^{2}+x^{2}-2x + 1=4\), \(2x^{2}-2x-3 = 0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 2\), \(b=-2\), \(c=-3\), \(x=\frac{2\pm\sqrt{4+24}}{4}=\frac{2\pm\sqrt{28}}{4}=\frac{2\pm2\sqrt{7}}{4}=\frac{1\pm\sqrt{7}}{2}\). Then \(y=x - 1=\frac{1\pm\sqrt{7}}{2}-1=\frac{-1\pm\sqrt{7}}{2}\). But for graph - recognition, we know the general shape: a circle centered at the origin with radius 2 and a line \(y=x - 1\) (which crosses the \(y\)-axis at \((0,-1)\) and has a slope of 1).

Answer:

The graph that shows a circle centered at the origin with radius 2 (\(x^{2}+y^{2}=4\)) and a line \(y=x - 1\) (with slope 1 and \(y\)-intercept \(-1\)) intersecting the circle at two points is the correct one.