If the points (2, 0), (0, 1), (4, 5)and (0, c) are concyclic, then the value of c, is

Options

A. 1
B. 14/3
C. 5
D. none of these

Answer

To find the value of \( c \) such that the points \( (2, 0) \), \( (0, 1) \), \( (4, 5) \), and \( (0, c) \) are concyclic, we will use the property that these points lie on the same circle. This can be determined using the determinant method for concyclic points.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
The points \( (x_1, y_1) = (2, 0) \), \( (x_2, y_2) = (0, 1) \), \( (x_3, y_3) = (4, 5) \), and \( (x_4, y_4) = (0, c) \) are concyclic if the following determinant is zero:

\[ \begin{vmatrix} x_1 & y_1 & x_1^2 + y_1^2 & 1 \\ x_2 & y_2 & x_2^2 + y_2^2 & 1 \\ x_3 & y_3 & x_3^2 + y_3^2 & 1 \\ x_4 & y_4 & x_4^2 + y_4^2 & 1 \end{vmatrix} = 0 \]

2. **Substituting the Points**:
Substitute the coordinates of the points into the determinant:

\[ \begin{vmatrix} 2 & 0 & 2^2 + 0^2 & 1 \\ 0 & 1 & 0^2 + 1^2 & 1 \\ 4 & 5 & 4^2 + 5^2 & 1 \\ 0 & c & 0^2 + c^2 & 1 \end{vmatrix} = 0 \]

This simplifies to:

\[ \begin{vmatrix} 2 & 0 & 4 & 1 \\ 0 & 1 & 1 & 1 \\ 4 & 5 & 41 & 1 \\ 0 & c & c^2 & 1 \end{vmatrix} = 0 \]

3. **Calculating the Determinant**:
We can expand this determinant using the first column:

\[ = 2 \begin{vmatrix} 1 & 1 & 1 \\ 5 & 41 & 1 \\ c & c^2 & 1 \end{vmatrix} - 0 + 0 - 0 \]

Now we need to calculate the determinant:

\[ = 2 \left( 1 \cdot \begin{vmatrix} 41 & 1 \\ c^2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 5 & 1 \\ c & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 5 & 41 \\ c & c^2 \end{vmatrix} \right) \]

Calculate each of the 2x2 determinants:

- \( \begin{vmatrix} 41 & 1 \\ c^2 & 1 \end{vmatrix} = 41 - c^2 \)
- \( \begin{vmatrix} 5 & 1 \\ c & 1 \end{vmatrix} = 5 - c \)
- \( \begin{vmatrix} 5 & 41 \\ c & c^2 \end{vmatrix} = 5c^2 - 41c \)

Putting it all together:

\[ = 2 \left( (41 - c^2) - (5 - c) + (5c^2 - 41c) \right) \]
\[ = 2 \left( 41 - c^2 - 5 + c + 5c^2 - 41c \right) \]
\[ = 2 \left( 36 + 4c^2 - 40c \right) \]
\[ = 8c^2 - 80c + 72 = 0 \]

4. **Solving the Quadratic Equation**:
We can simplify this equation:

\[ c^2 - 10c + 9 = 0 \]

Using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ c = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]
\[ = \frac{10 \pm \sqrt{100 - 36}}{2} \]
\[ = \frac{10 \pm \sqrt{64}}{2} \]
\[ = \frac{10 \pm 8}{2} \]

This gives us two possible values for \( c \):

\[ c = \frac{18}{2} = 9 \quad \text{and} \quad c = \frac{2}{2} = 1 \]

### Final Answer:
The possible values of \( c \) are \( c = 9 \) and \( c = 1 \).