If a point P has co-ordinates `(0, -2) and Q` is any point on the circle`x^2+y^2-5x-y+5=0,` then the maximum value of `(PQ)^2` is : (a) `(25+sqrt6)/2` (b) `14+5sqrt3` (c) `(47+10sqrt6)/2` (d) `8+5sqrt3`

Options

A. ` ( 25 + sqrt( 5))/( 2)`
B. `8+5 sqrt( 3)`
C. ` 14 + 5 sqrt( 3)`
D. `(47+ 10sqrt( 6))/(2)`

Answer

To solve the problem, we need to find the maximum value of \((PQ)^2\) where \(P\) has coordinates \((0, -2)\) and \(Q\) is any point on the circle given by the equation \(x^2 + y^2 - 5x - y + 5 = 0\).

### Step 1: Find the center and radius of the circle
We start by rewriting the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 - 5x - y + 5 = 0 \]

We can rearrange it as:

\[ x^2 - 5x + y^2 - y + 5 = 0 \]

Now, we complete the square for \(x\) and \(y\).

For \(x\):
\[ x^2 - 5x = (x - \frac{5}{2})^2 - \frac{25}{4} \]

For \(y\):
\[ y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4} \]

Substituting back into the equation gives:

\[ \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} + 5 = 0 \]

Combining the constants:

\[ \left(x - \frac{5}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{25}{4} + \frac{1}{4} - 5 \]

Calculating the right side:

\[ \frac{25 + 1 - 20}{4} = \frac{6}{4} = \frac{3}{2} \]

Thus, the center of the circle is \(\left(\frac{5}{2}, \frac{1}{2}\right)\) and the radius is \(\sqrt{\frac{3}{2}}\).

### Step 2: Find the distance from point \(P\) to the center of the circle
Next, we need to find the distance from point \(P(0, -2)\) to the center of the circle \(\left(\frac{5}{2}, \frac{1}{2}\right)\).

Using the distance formula:

\[ d = \sqrt{\left(\frac{5}{2} - 0\right)^2 + \left(\frac{1}{2} - (-2)\right)^2} \]

Calculating each part:

\[ d = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{1}{2} + 2\right)^2} = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{1}{2} + \frac{4}{2}\right)^2} = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2} \]

This simplifies to:

\[ d = \sqrt{2 \cdot \left(\frac{5}{2}\right)^2} = \sqrt{2 \cdot \frac{25}{4}} = \sqrt{\frac{50}{4}} = \frac{5\sqrt{2}}{2} \]

### Step 3: Calculate the maximum distance from \(P\) to any point \(Q\) on the circle
The maximum distance from point \(P\) to any point \(Q\) on the circle is the distance from \(P\) to the center plus the radius of the circle:

\[ \text{Max distance} = d + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} \]

### Step 4: Square the maximum distance
Now, we need to find \((PQ)^2\):

\[ (PQ)^2 = \left(\frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}}\right)^2 \]

Expanding this:

\[ = \left(\frac{5\sqrt{2}}{2}\right)^2 + 2 \cdot \frac{5\sqrt{2}}{2} \cdot \sqrt{\frac{3}{2}} + \left(\sqrt{\frac{3}{2}}\right)^2 \]

Calculating each term:

1. \(\left(\frac{5\sqrt{2}}{2}\right)^2 = \frac{25 \cdot 2}{4} = \frac{50}{4} = \frac{25}{2}\)
2. \(2 \cdot \frac{5\sqrt{2}}{2} \cdot \sqrt{\frac{3}{2}} = 5\sqrt{2} \cdot \sqrt{\frac{3}{2}} = 5\sqrt{3}\)
3. \(\left(\sqrt{\frac{3}{2}}\right)^2 = \frac{3}{2}\)

Combining these gives:

\[ (PQ)^2 = \frac{25}{2} + 5\sqrt{3} + \frac{3}{2} = \frac{28}{2} + 5\sqrt{3} = 14 + 5\sqrt{3} \]

### Final Answer
Thus, the maximum value of \((PQ)^2\) is:

\[ \boxed{14 + 5\sqrt{3}} \]