If the line passing through P=(8,3) meets the circle `S-=x^(2)+y^(2)-8x-10y+26=0` at A,B then PA.PB=

Options

A. 5
B. 10
C. 15
D. 25

Answer

To solve the problem, we need to find the product \( PA \cdot PB \) where \( P = (8, 3) \) and the points \( A \) and \( B \) are the intersection points of the line through \( P \) and the circle given by the equation:

\[ S: x^2 + y^2 - 8x - 10y + 26 = 0 \]

### Step 1: Rewrite the Circle Equation
First, we rewrite the circle equation in standard form by completing the square.

1. For \( x \):
\[ x^2 - 8x = (x - 4)^2 - 16 \]

2. For \( y \):
\[ y^2 - 10y = (y - 5)^2 - 25 \]

Putting it all together, we have:
\[ (x - 4)^2 - 16 + (y - 5)^2 - 25 + 26 = 0 \]
This simplifies to:
\[ (x - 4)^2 + (y - 5)^2 - 15 = 0 \]
Thus, the equation of the circle is:
\[ (x - 4)^2 + (y - 5)^2 = 15 \]

### Step 2: Identify the Center and Radius
From the standard form, we can identify the center and radius of the circle:
- Center \( C = (4, 5) \)
- Radius \( r = \sqrt{15} \)

### Step 3: Calculate \( S(8, 3) \)
Next, we substitute the coordinates of point \( P(8, 3) \) into the circle equation to find \( S(8, 3) \).

\[ S(8, 3) = 8^2 + 3^2 - 8 \cdot 8 - 10 \cdot 3 + 26 \]
Calculating each term:
- \( 8^2 = 64 \)
- \( 3^2 = 9 \)
- \( -8 \cdot 8 = -64 \)
- \( -10 \cdot 3 = -30 \)
- \( +26 = 26 \)

Now substituting these values:
\[ S(8, 3) = 64 + 9 - 64 - 30 + 26 \]
Simplifying this:
\[ S(8, 3) = 9 - 4 = 5 \]

### Step 4: Use the Power of a Point Theorem
According to the Power of a Point theorem, we have:
\[ PA \cdot PB = PT^2 \]
where \( PT \) is the distance from point \( P \) to the tangent point \( T \) on the circle.

Since we found \( S(8, 3) = 5 \), we can say:
\[ PT^2 = S(8, 3) = 5 \]

### Step 5: Conclusion
Thus, the product \( PA \cdot PB \) is:
\[ PA \cdot PB = 5 \]

### Final Answer
\[ PA \cdot PB = 5 \]