If a>0 and (a^2,2a) ,(1,-2) and (2,-1) never lie on a circle, then possible value of a is

a) 1

b) 2

c) 3

d) 4

To determine the possible values of \( a \), we analyze the given points:

1. \( (a^2, 2a) \)
2. \( (1, -2) \)
3. \( (2, -1) \)

### Step 1: Condition for Points to Lie on a Circle
For three points to lie on a circle, the determinant of the following matrix must be zero:

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

Since the problem states that these points **never** lie on a circle, this determinant should be **nonzero**.

### Step 2: Constructing the Determinant

Using the given points \( (a^2, 2a) \), \( (1, -2) \), and \( (2, -1) \), we set up the determinant:

\[
\begin{vmatrix}
a^2 & 2a & a^4 + 4a^2 & 1 \\
1 & -2 & 1 + 4 & 1 \\
2 & -1 & 4 + 1 & 1
\end{vmatrix} \neq 0
\]

### Step 3: Expanding the Determinant

Expanding along the first row:

\[
a^2 \begin{vmatrix} -2 & 5 & 1 \\ -1 & 5 & 1 \end{vmatrix}
- 2a \begin{vmatrix} 1 & 5 & 1 \\ 2 & 5 & 1 \end{vmatrix}
+ (a^4 + 4a^2) \begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix}
\]

#### Step 3.1: Compute the 2×2 Determinants

1. **First minor**:

\[
\begin{vmatrix} -2 & 5 \\ -1 & 5 \end{vmatrix} = (-2)(5) - (5)(-1) = -10 + 5 = -5
\]

2. **Second minor**:

\[
\begin{vmatrix} 1 & 5 \\ 2 & 5 \end{vmatrix} = (1)(5) - (5)(2) = 5 - 10 = -5
\]

3. **Third minor**:

\[
\begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (-2)(2) = -1 + 4 = 3
\]

### Step 4: Evaluating the Expression

\[
a^2 (-5) - 2a (-5) + (a^4 + 4a^2)(3) \neq 0
\]

\[
-5a^2 + 10a + 3a^4 + 12a^2 \neq 0
\]

\[
3a^4 + 7a^2 + 10a \neq 0
\]

\[
a(3a^3 + 7a + 10) \neq 0
\]

Since \( a > 0 \), the expression \( 3a^3 + 7a + 10 \neq 0 \) should hold for all valid values of \( a \). We check integer values from the given options.

For \( a = 1 \):

\[
3(1)^3 + 7(1) + 10 = 3 + 7 + 10 = 20 \neq 0
\]

For \( a = 2 \):

\[
3(2)^3 + 7(2) + 10 = 3(8) + 14 + 10 = 24 + 14 + 10 = 48 \neq 0
\]

For \( a = 3 \):

\[
3(3)^3 + 7(3) + 10 = 3(27) + 21 + 10 = 81 + 21 + 10 = 112 \neq 0
\]

For \( a = 4 \):

\[
3(4)^3 + 7(4) + 10 = 3(64) + 28 + 10 = 192 + 28 + 10 = 230 \neq 0
\]

Since the expression is **never zero** for any of the given values, **all options are possible**. However, the question asks for a single value. The smallest valid integer satisfying the inequality is **1**.

### Final Answer:
**a = 1**