Because $t=(x-3)^2$, only values with $t\ge 0$ can represent real solutions of the original system.

A positive value of $t$ gives two possible values of $x$, but $t=0$ gives only one. Use this fact to determine what kinds of roots the quadratic in $t$ must have.

Algebra is faster because the number of intersections depends on whether a value of $(x-3)^2$ is zero or positive. Desmos can verify the result after the substitution.

Graph $y=x^2-9x+25$ and restrict its domain to $x\ge0$. Here, the graph's $x$-coordinate represents the substituted value $t=(x-3)^2$.

Graph $y=r^2$ and use a positive slider for $r$. At $r=5$, the horizontal line intersects the restricted parabola at $x=0$ and $x=9$. The value $t=0$ corresponds to one original point, while $t=9$ corresponds to two original points.

Let $t=(x-3)^2$. Because it is a square, $t\ge 0$. From the parabola equation, $y-6=t-5$.

Substitute into the circle equation:

For every positive value of $t$, the equation $(x-3)^2=t$ has two real $x$-values, producing two points of intersection. For $t=0$, it has only one real $x$-value, $x=3$, producing one point.

To have exactly three solutions, the quadratic in $t$ must have $t=0$ as one root and one other positive root. Setting $t=0$ in $t^2-9t+25-r^2=0$ gives

Since $r$ is positive, $r=5$. The resulting equation is $t^2-9t=0$, whose roots are $0$ and $9$. These produce one and two intersection points, respectively, for a total of three.