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If the points $(2,0),(0,1),(4,5)$ and $(0, c)$ are concyclic, then the value of $c$ is
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Verified Answer
The correct answer is:
$\frac{14}{3}$
The equation of the circle passing through $(2,0),(0,1)$ and $(4,5)$ is
$$
3\left(x^{2}+y^{2}\right)-13 x-17 y+14=0
$$
This passes through $(0, c)$.
$$
\begin{array}{l}
\therefore 3 c^{2}-17 c+14=0 \\
\Rightarrow \quad c=1, \frac{14}{3}
\end{array}
$$
Since, $c=1$ is already there, for point $(0,1)$. Therefore, we take $c=\frac{14}{3}$.
$$
3\left(x^{2}+y^{2}\right)-13 x-17 y+14=0
$$
This passes through $(0, c)$.
$$
\begin{array}{l}
\therefore 3 c^{2}-17 c+14=0 \\
\Rightarrow \quad c=1, \frac{14}{3}
\end{array}
$$
Since, $c=1$ is already there, for point $(0,1)$. Therefore, we take $c=\frac{14}{3}$.
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