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

B

Explanation:

Answer 2

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

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.