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.
