Question

If Four distinct points $(2,3),(0,2),(4,5)$ and $(0,t)$ are concyclic, then $t^3+17$ is equal to

Answer 1

Let equation of the circle be $x^2+y^2+2gx+2fy+c=0$ .....(1)
Since it passes through the points $(2,3),(0,2),(4,5)$
We have $4g+6f+c=-13$ ....(2)
$4f+c=-4$ ....(3)
$8g+10f+c=-41$ ....(4)
Solving these equations, we get
$g=\frac{5}{2},f=-\frac{19}{2},c=34$
So, the equation of the circle is $x^2+y^2+5x-19y+34=0$
As this circle passes through the point $(0,t)$
$t^2-19t+34=0$
$\Rightarrow t=2$ or $t=17$.
But $t=2$ corresponds to the point $(0,2)$ which is different from $(0,t)$ Therefore $t=17$
So, $t^3+17=4930$