Because one equation is already solved for $y$, try substituting $y = x + 1$ into the other equation so you can solve for $x$.

After you substitute, you will get a quadratic equation in $x$. Expand $(x + 1)^2$, simplify, and factor the quadratic to find the two $x$-values.

Once you have the possible $x$-values, plug each back into $y = x + 1$ to get the corresponding $y$-values, then compare your points to the answer choices.

In Desmos, type x^2 + y^2 = 25 to graph the circle centered at the origin with radius 5.

On a new line in Desmos, type y = x + 1 to graph the line.

Click or tap on each point where the line and the circle cross. Desmos will display the coordinates of these intersection points; compare those coordinates to the answer choices.

The first equation $x^2 + y^2 = 25$ is a circle centered at the origin with radius $5$ (because $25 = 5^2$).

The second equation $y = x + 1$ is a straight line with slope $1$ and $y$-intercept $1$.

Their intersection points are the points that satisfy both equations at the same time.

Since $y = x + 1$, substitute $x + 1$ for $y$ in the circle equation:

Now you have an equation in terms of $x$ only.

Expand $(x + 1)^2$ and simplify:

Subtract $25$ from both sides:

Divide everything by $2$ to simplify:

This is a quadratic equation in $x$.

Factor the quadratic $x^2 + x - 12$:

Set each factor equal to zero:

• $x + 4 = 0$ gives $x = -4$.
• $x - 3 = 0$ gives $x = 3$.

So the possible $x$-values of the intersection points are $x = -4$ and $x = 3$.

Use $y = x + 1$ to find $y$ for each $x$:

• If $x = 3$, then $y = 3 + 1 = 4$, giving the point $(3, 4)$.
• If $x = -4$, then $y = -4 + 1 = -3$, giving the point $(-4, -3)$.

So, the two points of intersection are $(3, 4)$ and $(-4, -3)$, which matches choice A.