After you substitute $y = x - 1$, expand $(x-1)^2$ and combine like terms. You should get a quadratic equation in $x$; solve it with the quadratic formula.

For each $x$-value you find, plug it back into $y = x - 1$ to get $y$, then look for which answer choice uses one of those $(x,y)$ pairs.

In Desmos, type x^2 + y^2 = 10 to graph the circle centered at the origin with radius $\sqrt{10}$.

On a new line, type y = x - 1 to graph the line from the second equation.

Zoom in if needed and click on each point where the line intersects the circle. Desmos will display the coordinates of these intersection points; compare those coordinates to the answer choices to see which listed ordered pair matches one of them exactly.

Use $y=x-1$ from the second equation in the first equation.

Substitute into $x^2 + y^2 = 10$:

Now expand $(x-1)^2$:

Combine like terms:

Move 10 to the left to get a standard quadratic form:

Solve $2x^2 - 2x - 9 = 0$ using the quadratic formula. Here, $a=2$, $b=-2$, and $c=-9$.

Compute the discriminant:

So,

Since $76 = 4\cdot 19$, $\sqrt{76} = 2\sqrt{19}$, giving

So there are two possible $x$-values: $x = \frac{1+\sqrt{19}}{2}$ and $x = \frac{1-\sqrt{19}}{2}$.

Now use $y = x - 1$ to find $y$ for each $x$.

• If $x = \frac{1+\sqrt{19}}{2}$, then

• If $x = \frac{1-\sqrt{19}}{2}$, then

So the system has two solution points, each pairing one of these $x$-values with its matching $y$-value.

From the work above, the two solution pairs are:

• $\left(\frac{1+\sqrt{19}}{2},\,\frac{-1+\sqrt{19}}{2}\right)$, and
• $\left(\frac{1-\sqrt{19}}{2},\,\frac{-1-\sqrt{19}}{2}\right)$.

Now compare these to the answer choices:

• Choice A uses the same number for $x$ and $y$, so it does not satisfy $y=x-1$.
• Choice B mixes the $x$ with the minus sign and the $y$ with the plus sign, so $y$ is not equal to $x-1$.
• Choice C has the right pattern $y=x-1$ but uses $\sqrt{10}$ instead of $\sqrt{19}$, which does not satisfy $x^2 + y^2 = 10$.
• Choice D matches exactly one of the two solution pairs above.

Therefore, the correct answer is D) $\left(\frac{1+\sqrt{19}}{2},\,\frac{-1+\sqrt{19}}{2}\right)$.