Answer

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Hint: Use the general form of circle ${x^2} + {y^2} + 2gx + 2fy + d = 0$.

As we know,

The general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + d = 0..........\left( 1 \right)$

According to question all points $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)$ and $\left( {0,c} \right)$ are concyclic that means all points satisfy above general equation.

So, we put all points one by one and find the value of $g,f$ and $d$.

Now, put $\left( {2,0} \right)$ in (1) equation

$

4 + 0 + 4g + 0 + d = 0 \\

\Rightarrow 4 + 4g + d = 0...........\left( 2 \right) \\

$

Put $\left( {0,1} \right)$ in (1) equation

$

0 + 1 + 0 + 2f + d = 0 \\

\Rightarrow 1 + 2f + d = 0...........(3) \\

$

Put $(4,5)$ in (1) equation

$

16 + 25 + 8g + 10f + d = 0 \\

\Rightarrow 41 + 8g + 10f + d = 0...........\left( 4 \right) \\

$

Now, we have three equations of three variables.

So, we can easily get $g,f$ and $d$.

Solve $\left( 4 \right) - \left( 2 \right)$ and $\left( 4 \right) - \left( 3 \right)$ equation

$37 + 4g + 10f = 0..........\left( 5 \right)$ and

$

40 + 8g + 8f = 0 \\

\Rightarrow 5 + g + f = 0............\left( 6 \right) \\

\\

$

After solve (5) and (6) equation we get

$f = \frac{{ - 17}}{6}$ and $g = \frac{{ - 13}}{6}$

Now, put these values in (3) equation

$

\Rightarrow 1 - \frac{{34}}{6} + d = 0 \\

\Rightarrow d = \frac{{14}}{3} \\

$

So, equation of circle is ${x^2} + {y^2} - \frac{{13}}{3}x - \frac{{17}}{3}y + \frac{{14}}{3} = 0$

As we know $\left( {0,c} \right)$ also satisfy the equation of circle

$

\Rightarrow 0 + {c^2} - 0 - \frac{{17}}{3}c + \frac{{14}}{3} = 0 \\

\Rightarrow 3{c^2} - 17c + 14 = 0 \Rightarrow 3{c^2} - 3c - 14c + 14 = 0 \\

\Rightarrow 3c\left( {c - 1} \right) - 14\left( {c - 1} \right) = 0 \\

\Rightarrow \left( {c - 1} \right)\left( {3c - 14} \right) = 0 \\

\Rightarrow c = 1,\frac{{14}}{3} \\

$

So, the value of $c$ is $1,\frac{{14}}{3}$ .

Note: Whenever we face such types of problems we use some important points. Like we always use the general equation of circle and remember concyclic points always satisfy the general equation of circle, then after solving some equations we can easily get the required answer.

As we know,

The general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + d = 0..........\left( 1 \right)$

According to question all points $\left( {2,0} \right),\left( {0,1} \right),\left( {4,5} \right)$ and $\left( {0,c} \right)$ are concyclic that means all points satisfy above general equation.

So, we put all points one by one and find the value of $g,f$ and $d$.

Now, put $\left( {2,0} \right)$ in (1) equation

$

4 + 0 + 4g + 0 + d = 0 \\

\Rightarrow 4 + 4g + d = 0...........\left( 2 \right) \\

$

Put $\left( {0,1} \right)$ in (1) equation

$

0 + 1 + 0 + 2f + d = 0 \\

\Rightarrow 1 + 2f + d = 0...........(3) \\

$

Put $(4,5)$ in (1) equation

$

16 + 25 + 8g + 10f + d = 0 \\

\Rightarrow 41 + 8g + 10f + d = 0...........\left( 4 \right) \\

$

Now, we have three equations of three variables.

So, we can easily get $g,f$ and $d$.

Solve $\left( 4 \right) - \left( 2 \right)$ and $\left( 4 \right) - \left( 3 \right)$ equation

$37 + 4g + 10f = 0..........\left( 5 \right)$ and

$

40 + 8g + 8f = 0 \\

\Rightarrow 5 + g + f = 0............\left( 6 \right) \\

\\

$

After solve (5) and (6) equation we get

$f = \frac{{ - 17}}{6}$ and $g = \frac{{ - 13}}{6}$

Now, put these values in (3) equation

$

\Rightarrow 1 - \frac{{34}}{6} + d = 0 \\

\Rightarrow d = \frac{{14}}{3} \\

$

So, equation of circle is ${x^2} + {y^2} - \frac{{13}}{3}x - \frac{{17}}{3}y + \frac{{14}}{3} = 0$

As we know $\left( {0,c} \right)$ also satisfy the equation of circle

$

\Rightarrow 0 + {c^2} - 0 - \frac{{17}}{3}c + \frac{{14}}{3} = 0 \\

\Rightarrow 3{c^2} - 17c + 14 = 0 \Rightarrow 3{c^2} - 3c - 14c + 14 = 0 \\

\Rightarrow 3c\left( {c - 1} \right) - 14\left( {c - 1} \right) = 0 \\

\Rightarrow \left( {c - 1} \right)\left( {3c - 14} \right) = 0 \\

\Rightarrow c = 1,\frac{{14}}{3} \\

$

So, the value of $c$ is $1,\frac{{14}}{3}$ .

Note: Whenever we face such types of problems we use some important points. Like we always use the general equation of circle and remember concyclic points always satisfy the general equation of circle, then after solving some equations we can easily get the required answer.

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