Question

How do you find the exact solutions to the system `x^2+y^2=36` and `y=x+2`?

Answer 1

points:
`A(-1-root2(17),1-root2(17))` and `B(-1+root2(17),1+root2(17)`

Explanation:

If `y=x+2` this can be replaced inside the circumference so that its equation becomes
`x^2+(x+2)^2=36`
after developing the square we get the second degree equation
`2x^2+4x-32=0` that can further simplified into
`x^2+2x-16=0` that solved gives the solutions
`x_1=-1-root2(1+17)` and `x_2=-1+root2(1+16)` to which correspond the ordinates `y_1=1-root2(17)` and `y_2=1+root2(17)`