Statement I Tangents cannot be drawn from the point `(1,lamda)` to the circle `x^(2)+y^(2)+2x-4y=0`
Statement II `(1+1)^(2)+(lamda-2)^(2)lt1^(2)+2^(2)`

Options

A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I
B. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

To solve the problem, we need to analyze the given statements regarding the tangents from the point \((1, \lambda)\) to the circle defined by the equation \(x^2 + y^2 + 2x - 4y = 0\).

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 2x - 4y = 0 \]
We can rearrange it as:
\[ x^2 + 2x + y^2 - 4y = 0 \]

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]
2. For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Putting it all together:
\[ (x + 1)^2 - 1 + (y - 2)^2 - 4 = 0 \]
This simplifies to:
\[ (x + 1)^2 + (y - 2)^2 = 5 \]

### Step 3: Identify the Center and Radius
From the standard form \((x + 1)^2 + (y - 2)^2 = 5\), we can identify:
- Center \(C = (-1, 2)\)
- Radius \(r = \sqrt{5}\)

### Step 4: Calculate the Distance from Point to Center
Next, we need to find the distance from the point \((1, \lambda)\) to the center \((-1, 2)\). The distance \(d\) is given by:
\[ d = \sqrt{(1 - (-1))^2 + (\lambda - 2)^2} = \sqrt{(1 + 1)^2 + (\lambda - 2)^2} = \sqrt{4 + (\lambda - 2)^2} \]

### Step 5: Condition for Tangents
Tangents can be drawn from the point \((1, \lambda)\) to the circle if the distance \(d\) is greater than or equal to the radius \(r\):
\[ \sqrt{4 + (\lambda - 2)^2} \geq \sqrt{5} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ 4 + (\lambda - 2)^2 \geq 5 \]
This simplifies to:
\[ (\lambda - 2)^2 \geq 1 \]

### Step 7: Solve the Inequality
The inequality \((\lambda - 2)^2 \geq 1\) implies:
\[ \lambda - 2 \leq -1 \quad \text{or} \quad \lambda - 2 \geq 1 \]
This leads to:
\[ \lambda \leq 1 \quad \text{or} \quad \lambda \geq 3 \]

### Conclusion
For the point \((1, \lambda)\) to lie inside the circle (and thus no tangents can be drawn), \(\lambda\) must be in the interval:
\[ 1 < \lambda < 3 \]

### Final Statements
- **Statement I**: True, tangents cannot be drawn from the point \((1, \lambda)\) if \(1 < \lambda < 3\).
- **Statement II**: This statement is a mathematical representation of the condition derived above.