Respuesta :
Answer:
(-3,-4) and (4,3)
Explanation:
The equations of the circle and the line are given below:
[tex]\begin{gathered} x^2+y^2=25 \\ y=x-1 \end{gathered}[/tex]Substitute the second equation into the first:
[tex]\begin{gathered} x^2+y^2=25 \\ x^2+(x-1)^2=25 \\ x^2+(x-1)(x-1)=25 \\ x^2+x^2-x-x+1=25 \\ 2x^2-2x+1-25=0 \\ 2x^2-2x-24=0 \end{gathered}[/tex]Solve the equation for x:
[tex]\begin{gathered} 2(x^2-x-12)=0 \\ x^2-4x+3x-12=0 \\ x(x-4)+3(x-4)=0 \\ (x+3)(x-4)=0 \\ x=-3\text{ or }x=4 \end{gathered}[/tex]Next, find the corresponding values of y for each x:
When x=-3
[tex]\begin{gathered} y=-3-1 \\ y=-4 \\ \implies(-3,-4) \end{gathered}[/tex]When x=4
[tex]\begin{gathered} y=4-1 \\ y=3 \\ \implies(4,3) \end{gathered}[/tex]The points at which the circle and the line intersect are (-3,-4) and (4,3).
