Use substitution: replace $y$ in the circle equation with $x + t$ from the line equation. This will give you a single equation in terms of $x$ and $t$.

After substitution, you will get a quadratic equation in $x$. Recall that the discriminant $b^2 - 4ac$ tells you whether there are 0, 1, or 2 real solutions. What inequality should this discriminant satisfy for there to be exactly two distinct real solutions?

Once you find an inequality that $t$ must satisfy (it will involve $t^2$), rewrite it as an interval and list all integers in that interval. Then count how many there are.

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

Type y = x + t. Desmos will create a slider for the parameter t. This represents the family of lines with slope 1 and varying intercepts.

Move the t slider and watch how the line intersects the circle: for some values of t, there are no intersection points, for some there is exactly one (tangent), and for others there are two intersection points. Note the approximate range of t where there are two intersections.

To verify algebraically in Desmos, define D(t) = 200 - 4t^2. Then add a table of integer t values and evaluate D(t) for each. Identify for which integer t the discriminant D(t) is positive (these give two distinct real solutions), then count how many such integers there are.

The equation $x^2 + y^2 = 25$ represents a circle centered at the origin with radius $5$.

The equation $y = x + t$ represents a line with slope $1$ and $y$-intercept $t$.

For each real value of $t$, this line may:

• not intersect the circle (0 solutions),
• be tangent to the circle (1 solution), or
• intersect the circle in two points (2 solutions).

We want the values of $t$ that make the line intersect the circle in exactly two distinct points.

Substitute $y = x + t$ into the circle equation:

Now expand and simplify:

This is a quadratic equation in $x$ of the form $ax^2 + bx + c = 0$ with

• $a = 2$,
• $b = 2t$,
• $c = t^2 - 25$.

A quadratic equation $ax^2 + bx + c = 0$ has:

• two distinct real solutions if its discriminant $D = b^2 - 4ac$ is greater than 0.

Compute the discriminant here:

For two distinct real solutions, we need $D > 0$:

Divide both sides by $4$:

So $t$ must satisfy $t^2 < 50$.

From $t^2 < 50$, we can write this as an interval:

Since $\sqrt{50}$ is a bit more than $7$ (because $7^2 = 49$ and $8^2 = 64$), the integers $t$ that satisfy this inequality are:

Count them:

• From $-7$ to $-1$ is 7 integers,
• Then $0$ is 1 integer,
• From $1$ to $7$ is 7 integers.

Total: $7 + 1 + 7 = 15$ integer values of $t$ for which the system has exactly two distinct real solution pairs $(x, y)$. So the answer is 15.