If a line passes through the point `P(1,-2)` and cuts the `x^2+y^2-x-y= 0`at `A` and `B`, then the`maximum of `PA+PB` is

Options

A. `sqrt(26)`
B. 8
C. `sqrt(8)`
D. `2 sqrt(8)`

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - x - y = 0 \]
We can rewrite this equation in standard form by completing the square.

### Step 2: Completing the Square
Rearranging the equation:
\[ x^2 - x + y^2 - y = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{1}{2})^2 - \frac{1}{4} = 0 \]
This simplifies to:
\[ (x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2} \]
Thus, the center of the circle is \((\frac{1}{2}, \frac{1}{2})\) and the radius is \(\frac{1}{\sqrt{2}}\).

### Step 3: Parametric Form of the Line
We need to find the equation of a line passing through the point \(P(1, -2)\). The parametric form of a line can be expressed as:
\[ x = x_1 + r \cos \theta, \quad y = y_1 + r \sin \theta \]
Substituting \(P(1, -2)\):
\[ x = 1 + r \cos \theta, \quad y = -2 + r \sin \theta \]

### Step 4: Substitute into the Circle Equation
Substituting the parametric equations into the circle's equation:
\[ (1 + r \cos \theta)^2 + (-2 + r \sin \theta)^2 - (1 + r \cos \theta) - (-2 + r \sin \theta) = 0 \]
Expanding this:
\[ (1 + r \cos \theta)^2 + (-2 + r \sin \theta)^2 - 1 - r \cos \theta + 2 - r \sin \theta = 0 \]
This leads to:
\[ 1 + 2r \cos \theta + r^2 \cos^2 \theta + 4 - 4r \sin \theta + r^2 \sin^2 \theta - 1 - r \cos \theta + 2 - r \sin \theta = 0 \]
Combining like terms:
\[ r^2 + (2 \cos \theta - 4 \sin \theta - \cos \theta - \sin \theta) r + 6 = 0 \]
This simplifies to:
\[ r^2 + (cos \theta - 5 \sin \theta) r + 6 = 0 \]

### Step 5: Sum of Roots
The sum of the roots \(r_1 + r_2\) (which represent \(PA + PB\)) can be found using the formula:
\[ r_1 + r_2 = -\frac{b}{a} = -\left(\cos \theta - 5 \sin \theta\right) \]

### Step 6: Maximizing \(PA + PB\)
To maximize \(PA + PB\), we need to find the maximum value of:
\[ -\left(\cos \theta - 5 \sin \theta\right) \]
The maximum value of \(a \cos \theta + b \sin \theta\) is given by:
\[ \sqrt{a^2 + b^2} \]
Here, \(a = -1\) and \(b = 5\):
\[ \sqrt{(-1)^2 + (5)^2} = \sqrt{1 + 25} = \sqrt{26} \]

### Final Answer
Thus, the maximum value of \(PA + PB\) is:
\[ \sqrt{26} \]