Question

solve the system graphically.
x² + y² = 5
x - 9y = - 1

graph the system. choose the correct graph. all graphs are shown in a -12,12 by -6,6 window.

Explanation:

Step1: Analyze the first - equation

The equation $x^{2}+y^{2}=5$ represents a circle with center at the origin $(0,0)$ and radius $r = \sqrt{5}\approx2.24$.

Step2: Analyze the second - equation

Rewrite $x - 9y=-1$ as $x=9y - 1$. This is a linear equation in slope - intercept form (where if we write it in terms of $y$, $y=\frac{1}{9}x+\frac{1}{9}$), with slope $m=\frac{1}{9}$ and $y$ - intercept $b = \frac{1}{9}$.

Step3: Check the intersection points visually

We need to find the graph where a circle centered at the origin with radius $\sqrt{5}$ intersects a line with a positive - slope of $\frac{1}{9}$ and $y$ - intercept of $\frac{1}{9}$.

Since no specific graphs are shown here in a way that we can directly choose, we would look for a graph where a circle centered at $(0,0)$ with radius approximately $2.24$ intersects a line with a shallow positive slope and $y$ - intercept near the origin.

If we assume we have the correct visual understanding of the general shapes and positions of the circle and line:
We can also solve the system algebraically to find the exact intersection points. Substitute $x = 9y-1$ into $x^{2}+y^{2}=5$:
$$(9y - 1)^{2}+y^{2}=5$$
$$81y^{2}-18y + 1+y^{2}=5$$
$$82y^{2}-18y - 4 = 0$$
$$41y^{2}-9y - 2=0$$
Using the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ay^{2}+by + c = 0$, here $a = 41$, $b=-9$, $c=-2$
$$y=\frac{9\pm\sqrt{(-9)^{2}-4\times41\times(-2)}}{2\times41}=\frac{9\pm\sqrt{81 + 328}}{82}=\frac{9\pm\sqrt{409}}{82}$$
$$y_1=\frac{9+\sqrt{409}}{82}\approx\frac{9 + 20.22}{82}\approx0.356$$
$$y_2=\frac{9-\sqrt{409}}{82}\approx\frac{9 - 20.22}{82}\approx - 0.137$$
When $y_1\approx0.356$, $x_1=9y_1-1=9\times0.356-1=3.204 - 1 = 2.204$
When $y_2\approx - 0.137$, $x_2=9y_2-1=9\times(-0.137)-1=-1.233-1=-2.233$

We then look for the graph that has these intersection - like points.

Since we don't have the actual visual graphs to choose from in a way that we can directly point out the correct option, if we were to choose based on the general form:
The line $x - 9y=-1$ has a positive slope and the circle $x^{2}+y^{2}=5$ is centered at the origin. We would look for a graph where the line intersects the circle at two points, one in the first - quadrant and one in the third - quadrant.

Answer:

Without the actual visual content of the graphs A, B, C, D, we cannot give a definite answer. But the general approach is to look for a circle centered at the origin with radius $\sqrt{5}$ and a line with slope $\frac{1}{9}$ and $y$ - intercept $\frac{1}{9}$ intersecting at two points.