Question

How do you solve the system `x^2+y^2+8x+7=0` and `x^2+y^2+4x+4y-5=0` and `x^2+y^2=1`?

Answer 1

There are no points where all three equations intersect.

Explanation:

We have the following equations:

`E1: x^2+y^2+8x+7=0`
`E2: x^2+y^2+4x+4y-5=0`
`E3: x^2+y^2=1`

Let's substitute in E3 into E1 and E2, making the `x^2 and y^2` terms equal to 1:

`E1: 1+8x+7=0`
`E2: 1+4x+4y-5=0`

Now let's solve E1:

`E1: 8x+8=0`

`E1: 8x=-8`

`E1: x=-1`

Now let's solve E2:

`E2: 1+4x+4y-5=0`

`E2: 1+4(-1)+4y-5=0`

`E2: 1-4+4y-5=0`

`E2: -8+4y=0`

`E2: 4y=8`

`E2: y=2`

And now let's check our work by substituting into E3:

`E3: x^2+y^2=1`

`E3: (-1)^2+(2)^2=1`

`E3: 1+4!=1`

`E3: 5!=1`

So there is no solution in this system that satisfies all three equations.

We can see this in the following graphs:

This is the graph of `E1: x^2+y^2+8x+7=0`:

graph{x^2+y^2+8x+7=0 [-20, 20, -10, 10]}

This is the graph of `E2: x^2+y^2+4x+4y-5=0`:

graph{x^2+y^2+4x+4y-5=0 [-20, 20, -10, 10]}

This is the graph of `E3: x^2+y^2=1`:

graph{x^2+y^2=1 [-20, 20, -10, 10]}

As you can see, there are no points where all three graphs intersect.