Find the locus of the centre of the circle which cut the circles `x^2+y^2+4x-6y+9=0` and `x^2+y^2-4x+6y+4=0` orthogonally (a) `9x+10y-7=0` (b) `8x-12y+5=0` (c) `9x-10y+11=0` (d) `9x+10y+7=0`

Options

A. `9x+10y-7=0`
B. `x-y+2=0`
C. `9x-10y+11=0`
D. `9x+10y+7=0`

Explanation

To find the locus of the center of the circle that cuts the given circles orthogonally, we follow these steps:

### Step 1: Identify the equations of the circles
The given circles are:
1. \( C_1: x^2 + y^2 + 4x - 6y + 9 = 0 \)
2. \( C_2: x^2 + y^2 - 4x + 6y + 4 = 0 \)

### Step 2: Rewrite the equations in standard form
To find the centers and radii, we can rewrite the equations in the standard form of a circle \((x - h)^2 + (y - k)^2 = r^2\).

For \( C_1 \):
\[ x^2 + 4x + y^2 - 6y + 9 = 0 \]
Completing the square:
\[ (x^2 + 4x + 4) + (y^2 - 6y + 9) = 4 \]
\[ (x + 2)^2 + (y - 3)^2 = 4 \]
Thus, the center of \( C_1 \) is \( (-2, 3) \) and the radius is \( 2 \).

For \( C_2 \):
\[ x^2 - 4x + y^2 + 6y + 4 = 0 \]
Completing the square:
\[ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 9 \]
\[ (x - 2)^2 + (y + 3)^2 = 9 \]
Thus, the center of \( C_2 \) is \( (2, -3) \) and the radius is \( 3 \).

### Step 3: Use the orthogonality condition
The condition for two circles to intersect orthogonally is given by:
\[ s_1 - s_2 = 0 \]
Where \( s_1 \) and \( s_2 \) are the equations of the circles.

### Step 4: Set up the equation
We have:
\[ s_1 = x^2 + y^2 + 4x - 6y + 9 \]
\[ s_2 = x^2 + y^2 - 4x + 6y + 4 \]
Now, we find \( s_1 - s_2 \):
\[ s_1 - s_2 = (x^2 + y^2 + 4x - 6y + 9) - (x^2 + y^2 - 4x + 6y + 4) \]
This simplifies to:
\[ s_1 - s_2 = 4x + 4x - 6y - 6y + 9 - 4 = 8x - 12y + 5 = 0 \]

### Step 5: Write the locus equation
Thus, the locus of the center of the circle that cuts both circles orthogonally is:
\[ 8x - 12y + 5 = 0 \]

### Step 6: Identify the correct option
From the options provided, we see that:
(b) \( 8x - 12y + 5 = 0 \) is the correct answer.