The radical centre of circles represented by `S_(1)-=x^(2)+y^(2)-7x-6y-4=0,S_(2)-=x^(2)+y^(2)+10x+6y-4=0`

Options

A. `(-1,1)`
B. (1,-1)
C. (0,0)
D. none

Answer

To find the radical center of the circles represented by the equations \( S_1: x^2 + y^2 - 7x - 6y - 4 = 0 \) and \( S_2: x^2 + y^2 + 10x + 6y - 4 = 0 \), we will follow these steps:

### Step 1: Subtract the equations of the circles
We start by subtracting the second equation from the first:

\[ S_1 - S_2 = (x^2 + y^2 - 7x - 6y - 4) - (x^2 + y^2 + 10x + 6y - 4) = 0 \]

### Step 2: Simplify the equation
Now, we simplify the equation:

\[ S_1 - S_2 = -7x - 6y - 4 - (10x + 6y - 4) = -7x - 6y - 4 - 10x - 6y + 4 = -17x - 12y = 0 \]

### Step 3: Rearranging the equation
Rearranging gives us the equation of the radical axis:

\[ 17x + 12y = 0 \]

### Step 4: Finding the radical center
The radical center lies on the radical axis. To find the radical center, we can check the given options to see which point satisfies the equation of the radical axis.

#### Option 1: \((-1, 1)\)
Substituting \((-1, 1)\):

\[ 17(-1) + 12(1) = -17 + 12 = -5 \quad \text{(not equal to 0)} \]

#### Option 2: \((1, -1)\)
Substituting \((1, -1)\):

\[ 17(1) + 12(-1) = 17 - 12 = 5 \quad \text{(not equal to 0)} \]

#### Option 3: \((0, 0)\)
Substituting \((0, 0)\):

\[ 17(0) + 12(0) = 0 + 0 = 0 \quad \text{(equal to 0)} \]

### Conclusion
The radical center of the circles is at the point \((0, 0)\).

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