Find $p,q$ if $(2,3)$ is the circumcentre of the triangle whose sides are $2x+y=0$, $x-y-3=0$, and $x+py=q$
Is there some more sophisticated or easier way other than actually finding the point of intersection of the variable line with other two lines in terms of $p$ and $q$ and then $i)$solving equations that distance from all the three vertices is equal to the radius or $ii)$ using the fact that circumcentre is the point of intersection of perpendicular bisectors.
Since the lines $2x+y=0$ and $x-y-3=0$ intercect at $V_1=(1,-2)$, you know that the other vertices of the triangle belong to the circle centered at $(2,3)$ passing through $V_1$, that is, the circle$$(x-2)^2+(y-3)^2=26.$$Now, compute the other points at which this circle intersects the lines $2x+y=0$ and $x-y-3=0$. You will get $V_2=\left(\frac{-13}5,\frac{26}5\right)$ and $V_3=(7,4)$. So, $p$ and $q$ are such that the line $x+py=q$ is the line passing through $V_2$ and $V_3$. In other words, $p$ and $q$ are the solutions of the system$$\left\{\begin{array}{l}\frac{-13}5+\frac{26}5p=q\\7+4p=q,\end{array}\right.$$which are $p=8$ and $q=39$.