If the points $\left(2, 3\right)$, $\left(0, 2\right)$, $\left(4, 5\right)$ and $\left(0, t\right)$ are concyclic, then t=

Let $A\left(2, 3\right) B\left(0, 2\right) C\left(4, 5\right) and D\left(0, t\right)$

Equation of the circle with $\stackrel{\_\_\_}{AB}$ as a diameter is $x^2+y^2-2x-5y+6=0$

Equation to $\overleftrightarrow{AB}$ is $x-2y+4=0$

Equation of a circle passing through A, B is $x^2+y^2-2x-5y+6+\lambda \left(x-2y+4\right)=0$

This passes through $C\left(4, 5\right)\Rightarrow 16+25-8-25+$$\lambda \left(4-10+4\right)=0\Rightarrow 14-2\lambda =0\Rightarrow \lambda =7$

Equation of the circle passing through A, B & C is $x^2+y^2+5x-19y+34=0$

$D\left(0, t\right)$lies on this circle $\Rightarrow t^2-19t+34=0\Rightarrow \left(t-17\right)\left(t-2\right)=0\Rightarrow t=17 or 2.$