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:

$$\left|\begin{array}{cccc}{x}_1& y_1& x_1^2+y_1^2& 1\\ x_2& y_2& x_2^2+y_2^2& 1\\ x_3& y_3& x_3^2+y_3^2& 1\end{array}\right|=0$$

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:

$$\left|\begin{array}{cccc}{a}^2& 2a& a^4+4a^2& 1\\ 1& -2& 1+4& 1\\ 2& -1& 4+1& 1\end{array}\right|\ne 0$$

### Step 3: Expanding the Determinant

Expanding along the first row:

$$a^2\left|\begin{array}{ccc}-2& 5& 1\\ -1& 5& 1\end{array}\right|-2a\left|\begin{array}{ccc}1& 5& 1\\ 2& 5& 1\end{array}\right|+(a^4+4a^2)\left|\begin{array}{cc}1& -2\\ 2& -1\end{array}\right|$$

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

1. **First minor**:

$$\left|\begin{array}{cc}-2& 5\\ -1& 5\end{array}\right|=(-2)(5)-(5)(-1)=-10+5=-5$$

2. **Second minor**:

$$\left|\begin{array}{cc}1& 5\\ 2& 5\end{array}\right|=(1)(5)-(5)(2)=5-10=-5$$

3. **Third minor**:

$$\left|\begin{array}{cc}1& -2\\ 2& -1\end{array}\right|=(1)(-1)-(-2)(2)=-1+4=3$$

### Step 4: Evaluating the Expression

$$a^2(-5)-2a(-5)+(a^4+4a^2)(3)\ne 0$$

$$-5a^2+10a+3a^4+12a^2\ne 0$$

$$3a^4+7a^2+10a\ne 0$$

$$a(3a^3+7a+10)\ne 0$$

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

For $a=1$:

$$3(1{)}^3+7(1)+10=3+7+10=20\ne 0$$

For $a=2$:

$$3(2{)}^3+7(2)+10=3(8)+14+10=24+14+10=48\ne 0$$

For $a=3$:

$$3(3{)}^3+7(3)+10=3(27)+21+10=81+21+10=112\ne 0$$

For $a=4$:

$$3(4{)}^3+7(4)+10=3(64)+28+10=192+28+10=230\ne 0$$

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**

they should be collinear equation of line passing from (1,-2) and (2,-1) is
y+1=-1+2/2-1(x-2)
y+1=x-2
y=x-3 now if a^2,2a is also on the same line then
2a=a^2-3
a^2-2a-3=0
a=3