A ball moving around the circle `x^(2)+y^(2)-2x-4y-20=0` in anti-clockwise direction leaves it tangentially at the point P(-2,-2). After getting reflected from a straingt line, it passes through the centre of the circle. Find the equation of the straight line if its perpendicular distance from P is 5/2. You can assume that the angle of incidence is equal to the angle of reflection.

Answer

To solve the problem step by step, we will first rewrite the equation of the circle, find its center and radius, and then determine the equation of the straight line based on the given conditions.

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 4y - 20 = 0 \]
We will complete the square for both \(x\) and \(y\).

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

Substituting these into the circle equation gives:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 - 20 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 - 25 = 0 \]
Thus, we have:
\[ (x - 1)^2 + (y - 2)^2 = 25 \]

### Step 2: Identify the Center and Radius
From the equation \((x - 1)^2 + (y - 2)^2 = 25\):
- The center of the circle \((h, k)\) is \((1, 2)\).
- The radius \(r\) is \(\sqrt{25} = 5\).

### Step 3: Determine the Point of Tangency
The ball leaves the circle tangentially at point \(P(-2, -2)\).

### Step 4: Find the Equation of the Line
We know that the ball reflects off a straight line and then passes through the center of the circle \((1, 2)\).

Let the equation of the line be:
\[ Ax + By + C = 0 \]
The perpendicular distance \(d\) from point \(P(-2, -2)\) to the line is given by:
\[ d = \frac{|A(-2) + B(-2) + C|}{\sqrt{A^2 + B^2}} = \frac{5}{2} \]

### Step 5: Set Up the Distance Equation
This gives us the equation:
\[ | -2A - 2B + C | = \frac{5}{2} \sqrt{A^2 + B^2} \]

### Step 6: Use the Reflection Property
Since the ball reflects off the line and travels towards the center of the circle, the line must be perpendicular to the line connecting point \(P\) and the center \((1, 2)\).

The slope of the line connecting \(P\) and the center is:
\[ \text{slope} = \frac{2 - (-2)}{1 - (-2)} = \frac{4}{3} \]
Thus, the slope of the line (perpendicular) is:
\[ -\frac{3}{4} \]

### Step 7: Write the Equation of the Line
Using point-slope form, the equation of the line can be written as:
\[ y + 2 = -\frac{3}{4}(x + 2) \]
This simplifies to:
\[ y = -\frac{3}{4}x - \frac{3}{2} - 2 \]
\[ y = -\frac{3}{4}x - \frac{7}{2} \]

### Step 8: Convert to Standard Form
Multiplying through by 4 to eliminate the fraction:
\[ 4y = -3x - 14 \]
Rearranging gives:
\[ 3x + 4y + 14 = 0 \]

### Conclusion
The equation of the straight line is:
\[ 3x + 4y + 14 = 0 \]