The distance of the point (1, 2) from the radical axis of the circles `x^(2)+y^(2) +6x-16=0 and x^(2)+y^(2)-2x+6y-6=0` is

Options

A. `3//5`
B. `4//5`
C. `7//5`
D. none

Answer

To find the distance of the point (1, 2) from the radical axis of the given circles, we will follow these steps:

### Step 1: Write the equations of the circles
The equations of the circles are given as:
1. \( S_1: x^2 + y^2 + 6x - 16 = 0 \)
2. \( S_2: x^2 + y^2 - 2x + 6y - 6 = 0 \)

### Step 2: Find the radical axis
The radical axis can be found using the formula:
\[ S_1 - S_2 = 0 \]
Subtracting the equations of the circles:
\[ (x^2 + y^2 + 6x - 16) - (x^2 + y^2 - 2x + 6y - 6) = 0 \]
This simplifies to:
\[ 6x - (-2x) + 6y - 16 + 6 = 0 \]
\[ 8x - 6y - 10 = 0 \]
Thus, the equation of the radical axis is:
\[ 8x - 6y - 10 = 0 \]

### Step 3: Find the distance from the point (1, 2) to the line
The distance \( D \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by the formula:
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( 8x - 6y - 10 = 0 \), we have:
- \( A = 8 \)
- \( B = -6 \)
- \( C = -10 \)

Substituting the point \( (1, 2) \):
\[ D = \frac{|8(1) - 6(2) - 10|}{\sqrt{8^2 + (-6)^2}} \]
Calculating the numerator:
\[ D = \frac{|8 - 12 - 10|}{\sqrt{64 + 36}} = \frac{|-14|}{\sqrt{100}} = \frac{14}{10} = \frac{7}{5} \]

### Final Answer
The distance of the point (1, 2) from the radical axis of the circles is \( \frac{7}{5} \).

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