The distance of the point (1, 2) from the common chord of the circles `x^(2)+y^(2)-2x+3y-5=0, x^(2)+y^(2)+10x+8y-1=0` is

Options

A. 1
B. 2
C. 3
D. none

Answer

To find the distance of the point (1, 2) from the common chord of the circles given by the equations:

1. \( x^2 + y^2 - 2x + 3y - 5 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 10x + 8y - 1 = 0 \) (Circle 2)

we will follow these steps:

### Step 1: Write the equations of the circles in standard form
The equations of the circles are already given, but we can denote them as:
- Circle 1: \( S_1 = x^2 + y^2 - 2x + 3y - 5 = 0 \)
- Circle 2: \( S_2 = x^2 + y^2 + 10x + 8y - 1 = 0 \)

### Step 2: Find the equation of the common chord
To find the equation of the common chord, we subtract the second circle's equation from the first circle's equation:

\[ S_1 - S_2 = 0 \]

This gives us:
\[ (x^2 + y^2 - 2x + 3y - 5) - (x^2 + y^2 + 10x + 8y - 1) = 0 \]

Simplifying this:
\[ -2x + 3y - 5 - 10x - 8y + 1 = 0 \]
\[ -12x - 5y - 4 = 0 \]

Rearranging gives us the equation of the common chord:
\[ 12x + 5y + 4 = 0 \]

### Step 3: Use the distance formula from a point to a line
The distance \( d \) from a point \( (x_0, y_0) \) to a line given by the equation \( Ax + By + C = 0 \) is given by the formula:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our line \( 12x + 5y + 4 = 0 \):
- \( A = 12 \)
- \( B = 5 \)
- \( C = 4 \)

And for the point \( (1, 2) \):
- \( x_0 = 1 \)
- \( y_0 = 2 \)

### Step 4: Substitute the values into the distance formula
Substituting into the formula:

\[ d = \frac{|12(1) + 5(2) + 4|}{\sqrt{12^2 + 5^2}} \]

Calculating the numerator:
\[ = |12 + 10 + 4| = |26| = 26 \]

Calculating the denominator:
\[ \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

Thus, the distance is:
\[ d = \frac{26}{13} = 2 \]

### Final Answer
The distance of the point (1, 2) from the common chord of the circles is \( 2 \).

---