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. 15

Answer

To find the value of \( c \) such that the points \( (2, 0) \), \( (0, 1) \), \( (4, 5) \), and \( (0, c) \) are concyclic, we can use the property that four points are concyclic if the determinant of the following matrix is zero:

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

### Step 1: Set up the points
Let:
- \( (x_1, y_1) = (2, 0) \)
- \( (x_2, y_2) = (0, 1) \)
- \( (x_3, y_3) = (4, 5) \)
- \( (x_4, y_4) = (0, c) \)

### Step 2: Calculate \( x^2 + y^2 \) for each point
- For \( (2, 0) \): \( x_1^2 + y_1^2 = 2^2 + 0^2 = 4 \)
- For \( (0, 1) \): \( x_2^2 + y_2^2 = 0^2 + 1^2 = 1 \)
- For \( (4, 5) \): \( x_3^2 + y_3^2 = 4^2 + 5^2 = 16 + 25 = 41 \)
- For \( (0, c) \): \( x_4^2 + y_4^2 = 0^2 + c^2 = c^2 \)

### Step 3: Construct the determinant
The determinant becomes:

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

### Step 4: Calculate the determinant
Expanding this determinant:

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

Calculating each of these 3x3 determinants:

1. The first determinant:
\[ = 1 \begin{vmatrix} 5 & 1 \\ c & 1 \\ \end{vmatrix} - 1 \begin{vmatrix} 4 & 1 \\ 0 & 1 \\ \end{vmatrix} + 1 \begin{vmatrix} 4 & 5 \\ 0 & c \\ \end{vmatrix} \]
\[ = 5 - 4 + 4c = 1 + 4c \]

2. The second determinant:
\[ = 0 - 1 \cdot (4c^2 - 0) = -4c^2 \]

3. The third determinant:
\[ = 0 - 1 \cdot (4c - 0) = -4c \]

Putting it all together:
\[ 2(1 + 4c) + 4(-4c^2) - (-4c) = 0 \]
\[ 2 + 8c - 16c^2 + 4c = 0 \]
\[ -16c^2 + 12c + 2 = 0 \]

### Step 5: Solve the quadratic equation
Dividing through by -2:
\[ 8c^2 - 6c - 1 = 0 \]

Using the quadratic formula:
\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ = \frac{6 \pm \sqrt{36 + 32}}{16} = \frac{6 \pm \sqrt{68}}{16} = \frac{6 \pm 2\sqrt{17}}{16} = \frac{3 \pm \sqrt{17}}{8} \]

### Step 6: Find the value of \( c \)
The two possible values for \( c \) are:
\[ c = \frac{3 + \sqrt{17}}{8} \quad \text{and} \quad c = \frac{3 - \sqrt{17}}{8} \]

### Final Answer
The values of \( c \) that make the points concyclic are \( \frac{3 + \sqrt{17}}{8} \) and \( \frac{3 - \sqrt{17}}{8} \).