Question

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

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

\\) ?

Explanation:

Step1: Analyze the first equation

The equation \(x^{2}+y^{2} = 4\) represents a circle with center at the origin \((0,0)\) and radius \(r=\sqrt{4} = 2\) since the standard form of a circle is \((x - a)^{2}+(y - b)^{2}=r^{2}\) where \((a,b)\) is the center and \(r\) is the radius.

Step2: Analyze the second equation

The equation \(x - y=1\) can be rewritten in slope - intercept form (\(y=mx + c\)) as \(y=x - 1\). This is a straight line with a slope \(m = 1\) and a \(y\) - intercept \(c=- 1\).

Step3: Find the intersection points (optional for graph identification)

We can solve the system of equations. From \(x - y=1\), we have \(y=x - 1\). Substitute \(y=x - 1\) into \(x^{2}+y^{2}=4\):

\[

$$\begin{align*} x^{2}+(x - 1)^{2}&=4\\ x^{2}+x^{2}-2x + 1&=4\\ 2x^{2}-2x-3&=0 \end{align*}$$

\]

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 2\), \(b=-2\), \(c = - 3\)):

\[

$$\begin{align*} x&=\frac{2\pm\sqrt{(-2)^{2}-4\times2\times(-3)}}{2\times2}\\ &=\frac{2\pm\sqrt{4 + 24}}{4}\\ &=\frac{2\pm\sqrt{28}}{4}\\ &=\frac{2\pm2\sqrt{7}}{4}\\ &=\frac{1\pm\sqrt{7}}{2} \end{align*}$$

\]

When \(x=\frac{1+\sqrt{7}}{2}\), \(y=\frac{1+\sqrt{7}}{2}-1=\frac{-1 + \sqrt{7}}{2}\); when \(x=\frac{1-\sqrt{7}}{2}\), \(y=\frac{1-\sqrt{7}}{2}-1=\frac{-1-\sqrt{7}}{2}\)

The graph of the system should be a circle centered at the origin with radius 2 and a straight line with slope 1 and \(y\) - intercept - 1, intersecting the circle at two points.

Answer:

The graph should consist of a circle centered at \((0,0)\) with radius 2 and a straight line \(y = x-1\) intersecting the circle at two points. (To identify the specific graph from a set of options, look for a circle centered at the origin with radius 2 and a line with slope 1 and \(y\) - intercept - 1 that crosses the circle at two points.)