Given a circle described by the equation:
[tex]y^2+x^2=100[/tex]
and a function g(x) given by the table
[tex]\begin{center}
\begin{tabular}{ |c |c |c }
x & g(x) \\
-1 & -22 \\
0 & -20 \\
1 & -18
\end{tabular}
\end{center}[/tex]
The function g(x) describes a straight line with the equation:
[tex] \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \\ \\ \frac{y-(-22)}{x-(-1)} = \frac{-20-(-22)}{0-(-1)} \\ \\ \frac{y+22}{x+1} = \frac{-20+22}{0+1} = \frac{2}{1} \\ \\ y+22=2(x+1)=2x+2 \\ \\ y=2x-20[/tex]
To check if the circle and the line intersects, we substitute the equation of the line into the equation of the circle to see if we have a real solution.
i.e.
[tex]y^2+x^2=100 \\ \\ (2x-20)^2+x^2=100 \\ \\ 4x^2-80x+400+x^2=100 \\ \\ 5x^2-80x+300=0 \\ \\ x^2-16x+60=0 \\ \\ (x-6)(x-10)=0 \\ \\ x-6=0 \ or \ x-10=0 \\ \\ x=6 \ or \ x=10[/tex]
When x = 6, y = 2(6) - 20 = 12 - 20 = -8 and when x = 10, y = 2(10) - 20 = 20 - 20 = 0
Therefore, the circle and the line intersect at the points (6, -8) and (10, 0).
