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question mark

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

Pawan joshi , 6 Years ago
Grade 12th pass
anser 2 Answers
Saurabh Koranglekar

Last Activity: 6 Years ago

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

1. (a2,2a)
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:

|x1y1x12+y121x2y2x22+y221x3y3x32+y321|=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 (a2,2a), (1,2), and (2,1), we set up the determinant:

|a22aa4+4a21121+41214+11|0

### Step 3: Expanding the Determinant

Expanding along the first row:

a2|251151|2a|151251|+(a4+4a2)|1221|

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

1. **First minor**:

|2515|=(2)(5)(5)(1)=10+5=5

2. **Second minor**:

|1525|=(1)(5)(5)(2)=510=5

3. **Third minor**:

|1221|=(1)(1)(2)(2)=1+4=3

### Step 4: Evaluating the Expression

a2(5)2a(5)+(a4+4a2)(3)0

5a2+10a+3a4+12a20

3a4+7a2+10a0

a(3a3+7a+10)0

Since a>0, the expression 3a3+7a+100 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=200

For a=2:

3(2)3+7(2)+10=3(8)+14+10=24+14+10=480

For a=3:

3(3)3+7(3)+10=3(27)+21+10=81+21+10=1120

For a=4:

3(4)3+7(4)+10=3(64)+28+10=192+28+10=2300

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**

Deepak Kumar Shringi

Last Activity: 6 Years ago

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

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