Answer:
Yes, at positive x coordinates
Step-by-step explanation:
Find the equation of g(x)
Given ordered pairs of g(x): (-1, -22) (0, -20) (1, -18)
[tex]\sf let\:(x_1,y_1)=(0,-20)[/tex]
[tex]\sf let\:(x_2,y_2)=(1,-18)[/tex]
[tex]\sf slope\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-18-(-20)}{1-0}=2[/tex]
Point-slope form of linear function: [tex]\sf y-y_1=m(x-x_1)[/tex]
[tex]\implies \sf y-(-20)=2(x-0)[/tex]
[tex]\implies \sf y=2x-20[/tex]
Substitute the equation of g(x) into the equation of the circle and solve for x
Given equation: [tex]y^2+x^2=100[/tex]
[tex]\implies (2x-20)^2+x^2=100[/tex]
[tex]\implies 4x^2-80x+400+x^2=100[/tex]
[tex]\implies 5x^2-80x+300=0[/tex]
[tex]\implies x^2-16x+60=0[/tex]
[tex]\implies x^2-10x-6x+60=0[/tex]
[tex]\implies x(x-10)-6(x-10)=0[/tex]
[tex]\implies (x-6)(x-10)=0[/tex]
Therefore:
[tex](x-6)=0 \implies x=6[/tex]
[tex](x-10)=0 \implies x=10[/tex]
So the linear function g(x) will intersect the equation of the circle at positive x coordinates.
