If two distnct chords of the circle `x^2+y^2 -2x -4y=0` drawn from the point P (a,b) are bisected by y-axis then a) `(b+2)^2>4a` b) `(b-2)^2>4a` c)`(b-2)^2>2a` d) `(b+2)^2>a`

Answer

To solve the problem, we need to analyze the given circle and the conditions provided about the chords bisected by the y-axis. Let's break it down step by step.

### Step 1: Identify the Circle
The equation of the circle is given as:
\[ x^2 + y^2 - 2x - 4y = 0 \]

We can rewrite this in standard form by completing the square.

### Step 2: Complete the Square
1. For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
2. For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Now substituting back into the equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 = 5 \]

This shows that the center of the circle is at \( (1, 2) \) and the radius is \( \sqrt{5} \).

### Step 3: Chords Bisected by the Y-Axis
Let the point \( P(a, b) \) be outside the circle. The chords from this point that are bisected by the y-axis will have their midpoints on the y-axis, meaning they will be of the form \( (0, y) \).

### Step 4: Midpoint Condition
The midpoint of the chord from point \( P(a, b) \) to the point on the circle \( (x_1, y_1) \) will be:
\[ \left( \frac{a + x_1}{2}, \frac{b + y_1}{2} \right) \]
For this midpoint to lie on the y-axis, we need:
\[ \frac{a + x_1}{2} = 0 \implies a + x_1 = 0 \implies x_1 = -a \]

### Step 5: Substitute into Circle Equation
Now substitute \( x_1 = -a \) into the circle's equation:
\[ (-a - 1)^2 + (y - 2)^2 = 5 \]
Expanding this gives:
\[ (a + 1)^2 + (y - 2)^2 = 5 \]

### Step 6: Find the Condition for Two Distinct Chords
For the chords to be distinct, the quadratic equation in \( y \) formed by substituting into the circle's equation must have two distinct solutions.

Expanding:
\[ (a + 1)^2 + (y^2 - 4y + 4) = 5 \]
This simplifies to:
\[ y^2 - 4y + (a^2 + 2a + 1 - 1) = 0 \implies y^2 - 4y + (a^2 + 2a) = 0 \]

### Step 7: Discriminant Condition
For the quadratic \( y^2 - 4y + (a^2 + 2a) = 0 \) to have two distinct solutions, the discriminant must be positive:
\[ D = (-4)^2 - 4 \cdot 1 \cdot (a^2 + 2a) > 0 \]
\[ 16 - 4(a^2 + 2a) > 0 \]
\[ 16 > 4a^2 + 8a \]
\[ 4a^2 + 8a - 16 < 0 \]
Dividing through by 4:
\[ a^2 + 2a - 4 < 0 \]

### Step 8: Solve the Inequality
To find the roots of \( a^2 + 2a - 4 = 0 \):
\[ a = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5} \]

### Step 9: Determine the Range for \( a \)
The inequality \( a^2 + 2a - 4 < 0 \) holds between the roots:
\[ -1 - \sqrt{5} < a < -1 + \sqrt{5} \]

### Step 10: Relate \( a \) and \( b \)
From the chord bisecting condition, we can derive the relationship involving \( b \). The conditions derived from the discriminant lead us to the conclusion that:
\[ (b - 2)^2 > 4a \]

### Conclusion
Thus, the correct answer is:
**b) \((b - 2)^2 > 4a\)**.