Answer
$(4,3),(-3,-4)$
Work Step by Step
We substitute $y=x-1$ in the equation $x^2+y^2=25$.
$$
\begin{aligned}
x^2+(x-1)^2 & =25 \\
x^2+x^2-2 x+1 & =25 \\
2 x^2-2 x-24 & =0 \\
x^2-x-12 & =0 \\
(x-4)(x+3) & =0
\end{aligned}
$$
$x=4 \quad$ and $\quad y=4-1=3 \quad$ or $x=-3 \quad$ and $\quad y=-3-1=-4$
The points of intersection are $(4,3),(-3,-4)$.
