Christopher is analyzing a circle, y^2 + x^2 = 121, and a linear function g(x). Will they intersect? A. Yes, at the positive x coordinates. B. Yes, at negative x coordinates. C. Yes, at the negative and positive x coordinates. D. No, they will not intersect

Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12   (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case, 
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.