To solve the exercise, first replace the value of y of the line in the equation of the circle, like this
[tex]\begin{gathered} y=x-3 \\ (x-1)^2+y^2=4 \end{gathered}[/tex][tex](x-1)^2+(x-3)^2=4[/tex]Solving for x you have
[tex]\begin{gathered} (x-1)(x-1)+(x-3)(x-3)=4 \\ x^2-x-x+1+x^2-3x-3x+9=4 \end{gathered}[/tex]Subtract 4 from both of the equation and operate like terms
[tex]\begin{gathered} x^2-x-x+1+x^2-3x-3x+9-4=4-4 \\ 2x^2-8x+6=0 \end{gathered}[/tex]Now you can use the general formula for quadratic equations, that is
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In this case
[tex]\begin{gathered} 2x^2-8x+6=0 \\ a=2 \\ b=-8 \\ c=6 \end{gathered}[/tex]So, you have
[tex]\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(2)(6)}}{2(2)} \\ x=\frac{8\pm\sqrt[]{64-48}}{4} \\ x=\frac{8\pm\sqrt[]{16}}{4} \\ x=\frac{8\pm4}{4} \end{gathered}[/tex]Then the solutions of the quadratic equation will be
[tex]\begin{gathered} x_1=\frac{8+4}{4} \\ x_1=\frac{12}{4} \\ x_1=3 \end{gathered}[/tex][tex]\begin{gathered} x_2=\frac{8-4}{4} \\ x_2=\frac{4}{4} \\ x_2=1 \end{gathered}[/tex]Finally, to get the y-coordinate of the intersection points between the line and the circle, replace the values of x found in any of the initial equations, for example in the equation of the line
[tex]\begin{gathered} y_1=x_1-3 \\ y_1=3-3 \\ y_1=0 \end{gathered}[/tex][tex]\begin{gathered} y_2=x_2-3 \\ y_2=1-3 \\ y_2=-2 \end{gathered}[/tex]Therefore, the points where the line and the circle intersect are
[tex](3,0)\text{ and }(1,-2)[/tex]And the correct answers are E. and C.
