Question

Which graph represents the solution of the system StartLayout Enlarged left-brace 1st row x squared + y squared = 4 2nd row x minus y = 1 EndLayout?
On a coordinate plane, a graph of a circle and a straight line are shown. The circle has a center at (0, 0) and a radius of 4 units. The line has a negative slope and goes through (negative 3, 4), (0, 1), and (1, 0).
On a coordinate plane, a graph of a circle and a straight line are shown. The circle has a center at (0, 0) and a radius of 2 units. The line has a positive slope and goes through (negative 3, negative 4), (0, negative 1), and (1, 0).
On a coordinate plane, a graph of a circle and a straight line are shown. The circle has a center at (0, 0) and a radius of 4 units. The line has a positive slope and goes through (negative 3, negative 4), (0, negative 1), and (1, 0).
On a coordinate plane, a graph of a circle and a straight line are shown. The circle has a center at (0, 0) and a radius of 2 units. The line has a negative slope and goes through (negative 3, 4), (0, 1), and (1, 0).

Answer 1

Answer:

The answer is B. on edge

Step-by-step explanation:

I just took it, the positive linear line going through the smaller circle

Answer 2

Answer:

The solutions of the system are $(1.8,0.8)$ and $(-0.8, -1.8)$.

Step-by-step explanation:

The given system is

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

This is nonlinear system of equations, notice that it's about the intersection between a circle and a line, which can be in one point (tangent) or in two points (secant).

Let's isolate $x$ in the second equation, and then we replace that expression into the first equation

$x=y+1$

$x^{2} +y^{2} \\(y+1)^{2} +y^{2} =4\\y^{2}+ 2y+1+y^{2}=4 \\2y^{2} +2y-3=0$

Using a calculator, the solutions are

$y_{1} \approx 0.8\\y_{2} \approx -1.8$

Now, we use these values, to find their pairs.

$x=y+1\\x=0.8+1 \approx 1.8$

$x=y+1\\x=-1.8+1 \approx -0.8$

Therefore, the solutions of the system are $(1.8,0.8)$ and $(-0.8, -1.8)$.

The image attached shows the solutions graphically.