The points on the line segment each satisfy an equation and two inequalities,
$$ x=3 \qquad y>1 \qquad y<5. $$
The points on the circle each satisfy the equation,
$$ (x-5)^2+(y-4)^2=9. $$
An intersection of these curves will be an ordered pair, $(x,y)$, which solves both equations and satisfies the inequalities.
Since every point on the vertical line segment has the x-coordinate $x=3$ we will start by substituting $x=3$ into the equation of the circle.
$$ (3-5)^2+(y-4)^2=9 $$
$$ 4 + (y-4)^2 = 9 $$
$$ (y-4)^2=5 $$
$$ y-4=\pm\sqrt5 $$
$$ y=4\pm\sqrt5 $$
$$ y \approx 4 \pm 2.23 $$
$$ y\approx 1.77, 6.23$$
This gives us two ordered pairs $(3,1.77)$ and $(3,6.23)$. These both solve the equations for the circle and the line segment, but only one of them satisfies the inequalities. Since $1<y<5$, only the point $(3,1.77)$ is a valid point of intersection.
Expanding the determinant in terms of $3\times 3$ determinants, using the top row, we obtain an equation $A(x^2+y^2)+B x +C y +D =0$ with constants $A,B,C,D.$ The solution set to an equation of this form is (i) empty,or (ii) contains one point, or (iii) is a line,or (iv) is a circle, or (v) is the whole plane. Since the top row co-incides with one of the other rows when $(x,y)=(x_i,y_i)$ for $i\in \{1,2,3\},$ the $3$ given points belong to the solution set, so we can rule out (i) and (ii). And (iii) or (v) each require $A=0,$ but the co-efficient $A\ne 0$ because the $3$ given points are not co-linear. This leaves (iv).And as already stated, the $3$ given points belong to the solution set, that is, they lie on the circle.
Best Answer
The points $(x,y)$ on the line segment that joins $(x_1,y_1)$ and $(x_2,y_2)$ can be represented parametrically by $$x=tx_1+(1-t)x_2, \qquad y=ty_1+(1-t)y_2,$$ where $0\le t\le 1$. Substitute in the equation of the circle, solve the resulting quadratic for $t$. If $0\le t\le 1$ we have an intersection point, otherwise we don't. The value(s) of $t$ between $0$ and $1$ (if any) determine the intersection point(s).
If we want a simple yes/no answer, we can use the coefficients of the quadratic in $t$ to determine the answer without taking any square roots.