After you substitute $x+3$ for $y$ in the second equation, you will get an equation with only $x$. Expand the square, combine like terms, and see what kind of equation you get.

Once you have the quadratic in $x$, use the quadratic formula. You should end up with two possible $x$ values—think about which one is positive, as the question specifically asks for the positive value of $x$.

In Desmos, enter the equation y = x + 3 to graph the line from the first equation.

On a new line, enter x^2 + y^2 = 25 to graph the circle with radius 5 centered at the origin.

Look for the intersection points of the line and the circle. Click on each intersection, and note the $x$-coordinates; the one that is greater than 0 is the positive solution for $x$ that corresponds to the correct answer choice.

We are given

Since $y = x+3$, substitute $x+3$ for $y$ in the second equation so everything is in terms of $x$:

Now expand $(x+3)^2$ and combine like terms:

So the equation becomes

Combine like terms:

Subtract 25 from both sides:

Divide by 2 to simplify:

For the quadratic $x^2 + 3x - 8 = 0$, use the quadratic formula

with $a=1$, $b=3$, and $c=-8$.

So

Simplify the expression under the square root:

so the two solutions are

That is,

• $x = \dfrac{-3 + \sqrt{41}}{2}$
• $x = \dfrac{-3 - \sqrt{41}}{2}$

The question asks for the positive value of $x$. Since $\sqrt{41} \approx 6.4$,

• $-3 + \sqrt{41} \approx 3.4$ (positive)
• $-3 - \sqrt{41} \approx -9.4$ (negative)

So the positive solution is

which corresponds to choice D.