Question

If $(2,0)$, $(0,1)$, $(4,5)$ and $(0,c)$ are concyclic, then find $c$.

Answer 1

Let the center of circle passing through the points be $(h,k)$

The distance from center to all four points are equal

We have ${(h-2)}^2+{(k-0)}^2={(h-0)}^2+{(k-1)}^2$

$\Rightarrow 4h-2k=3$

We have ${(h-2)}^2+{(k-0)}^2={(h-4)}^2+{(k-5)}^2$

$\Rightarrow 4h+10k=37$

By solving those two equations , we get $h=\frac{17}{6}$ and $k=\frac{25}{6}$

The radius of circle is $\sqrt{\frac{25}{36}+\frac{625}{36}}=\sqrt{\frac{650}{36}}=\frac{5}{6}\sqrt{26}$

We have ${(\frac{17}{6}-0)}^2+{(\frac{25}{6}-c)}^2=\frac{650}{36}$

$\Rightarrow {(\frac{25}{6}-c)}^2=\frac{361}{36}$

$\Rightarrow \frac{25}{6}-c=\frac{19}{6},-\frac{19}{6}$

$\Rightarrow c=1,\frac{22}{3}$

Therefore the value of $c$ is $\frac{22}{3}$