Equation

$\displaystyle \begin{aligned} x^2 + y^2 + 13 & = 4x + 6y \\ \left(x^2 - 4x\right) + \left(y^2 - 6y\right) & = -13 & \small \color{#3D99F6} \text{Complete the square} \\ \left(x^2 - 4x \ {\color{#20A900}{+ \ 4}}\right) {\color{#20A900}{+}} \left(y^2 - 6y \ {\color{#20A900}{+ \ 9}}\right) & = -13 \ {\color{#20A900}{+ \ 13}} \\ (x - 2)^2 + (y - 3)^2 & = 0 \end{aligned}$

As Rajdeep points out in the comments, the above is the standard equation of a degenerate circle with radius $r = 0$ with center $(2, 3)$ . This point is the only one in the circle that satisfies the equation, and we conclude that there is one unique solution $(x, y) = (2, 3)$ to this equation.

$\begin{aligned} x^2 + y^2 + 13 & = 4x+6y \\ x^2 - 4x + 4 + y^2 - 6y + 9 & = 0 \\ (x-2)^2 + (y-3)^2 & = 0 \end{aligned}$

We note that the $RHS=0$ . Since $(x-2)^2 \ge 0$ and $(y-3)^2 \ge 0$ , for the $LHS = 0$ , both $(x-2)^2$ and $(y-3)^2$ must be equal to 0. That is when $x=2$ and $y=3$ and there is only $\boxed{1}$ solution.

$\begin{aligned} x^2+y^2+13 & =4x+6y \\ x^2-4x+y^2-6y+13 & =0 \\ (x-2)^2-4+(y-3)^2-9+13 & =0 \\ (x-2)^2+(y-3)^2-13+13 & =0 \\ (x-2)^2+(y-3)^2 & =0 \end{aligned}$

Since for each $r$ real number, $r^2\geq0$ , $x-2=0 \Rightarrow x=2$ and $y-3=0 \Rightarrow y=3$ So the only solution for the equation is: $2^2+3^3+13=4*2+6*3$