To get the point of intersection, you must equate the two equations and solve for x and y.
[tex](x-2)^{2} + (y+3)^{2} = 18 -> equation1 \\
y=-2x-2 -> equation 2\\
x^{2}-4x+4 + (-2x-2+3)^{2} = 18 \\
x^{2}-4x+4 + (-2x+1)^{2} = 18 \\
x^{2}-4x+4 + 4x^{2} -4x+1=18\\
5x^{2}-8x+5=18\\
5x^{2}-8x-13=0\\
(5x-13)(x+1)=0\\
5x-13 = 0 | x+1 = 0\\
x=\frac{13}{5} | x=-1\\
[/tex]
[tex]y=-2x-2 \\
y = -2(\frac{13}{5})-2 | y=-2(-1)-2 \\
y=\frac{-26}{5}-2 | y=2-2 \\
y=\frac{-36}{5}|y=0 \\
(\frac{13}{5},\frac{-36}{5}) and (-1,0)[/tex]
