Use the fact that the center lies on the line $y = -x + 10$. Substitute the $y$-coordinate you found (in terms of $r$) into this equation to express the center's $x$-coordinate in terms of $r$.

Once you can write the center as $(10 - r, r)$, apply the distance formula between this point and $(6,4)$, set that distance equal to $r$, and solve the resulting quadratic equation. Then use the condition that the center is in the first quadrant to decide which of the two possible $r$-values is valid.

From the algebra, the radius $r$ must satisfy $r^2 = 2(4 - r)^2$. In Desmos, graph the two functions by typing y = x^2 in one line and y = 2(4 - x)^2 in another; the $x$-values where these graphs intersect are the solutions to this equation.

Click on each intersection point of the two parabolas and note the $x$-coordinates; these are the two positive values that both satisfy the distance equation and are candidates for the radius.

In a new expression line, type 10 - x and evaluate it at each intersection $x$-value. Only the $x$-value for which 10 - x is positive makes the center $(10 - x, x)$ lie in the first quadrant; that $x$-value is the circle's radius.

Let the radius be $r$. Because the circle is tangent to the $x$-axis and its center is in the first quadrant, the vertical distance from the center to the $x$-axis is $r$, so the center's $y$-coordinate is $r$.

The center also lies on the line $y = -x + 10$. Substituting $y = r$ gives

Solving for $x$ gives $x = 10 - r$, so we can write the center as $(10 - r, r)$.

Because the point $(6,4)$ lies on the circle, the distance from the center $(10 - r, r)$ to $(6,4)$ must equal the radius $r$.

Using the distance formula,

Simplify $10 - r - 6$ to $4 - r$:

Since $(4 - r)^2 = (r - 4)^2$, this becomes

Now expand $(4 - r)^2$:

Substitute into $2(4 - r)^2 = r^2$:

Distribute the 2:

Move all terms to one side to get a quadratic equation:

Apply the quadratic formula to $r^2 - 16r + 32 = 0$.

The discriminant is

So

Since $\sqrt{128} = \sqrt{64\cdot 2} = 8\sqrt{2}$, this simplifies to

The center is $(10 - r, r)$ and must be in the first quadrant, so both coordinates must be positive. If $r = 8 + 4\sqrt{2}$, then $10 - r$ is negative, so that radius is not valid. If $r = 8 - 4\sqrt{2}$, both $r$ and $10 - r$ are positive, so the radius of the circle is $8 - 4\sqrt{2}$.