Find the equation of the circle which passes through the origin and intersects each of the following circles orthogonally.
`x^2+y^2-4x-6y-3=0`
`x^2+y^2-8y+12=0`

Answer

To find the equation of the circle that passes through the origin and intersects the given circles orthogonally, we will follow these steps:

### Step 1: Identify the equations of the given circles.
The equations of the circles are:
1. \( C_1: x^2 + y^2 - 4x - 6y - 3 = 0 \)
2. \( C_2: x^2 + y^2 - 8y + 12 = 0 \)

### Step 2: Rewrite the equations in standard form.
We can rewrite the equations of the circles in standard form by completing the square.

For circle \( C_1 \):
\[ x^2 - 4x + y^2 - 6y = 3 \]
Completing the square:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 3 \]
\[ (x - 2)^2 + (y - 3)^2 = 16 \]
This represents a circle with center \( (2, 3) \) and radius \( 4 \).

For circle \( C_2 \):
\[ x^2 + (y^2 - 8y) = -12 \]
Completing the square:
\[ x^2 + (y - 4)^2 - 16 = -12 \]
\[ x^2 + (y - 4)^2 = 4 \]
This represents a circle with center \( (0, 4) \) and radius \( 2 \).

### Step 3: Set up the equation of the required circle.
Let the equation of the required circle be:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Since it passes through the origin, we have \( c = 0 \):
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

### Step 4: Use the orthogonality condition.
For two circles to intersect orthogonally, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

#### For circle \( C_1 \):
- \( g_1 = -2 \), \( f_1 = -3 \), \( c_1 = -3 \)

#### For the required circle:
- \( g_2 = g \), \( f_2 = f \), \( c_2 = 0 \)

Applying the orthogonality condition for \( C_1 \) and the required circle:
\[ 2(-2)(g) + 2(-3)(f) = -3 + 0 \]
\[ -4g - 6f = -3 \quad \text{(Equation 1)} \]

#### For circle \( C_2 \):
- \( g_1 = 0 \), \( f_1 = -4 \), \( c_1 = 12 \)

Applying the orthogonality condition for \( C_2 \) and the required circle:
\[ 2(0)(g) + 2(-4)(f) = 12 + 0 \]
\[ -8f = 12 \quad \Rightarrow \quad f = -\frac{3}{2} \quad \text{(Equation 2)} \]

### Step 5: Substitute \( f \) into Equation 1.
Substituting \( f = -\frac{3}{2} \) into Equation 1:
\[ -4g - 6\left(-\frac{3}{2}\right) = -3 \]
\[ -4g + 9 = -3 \]
\[ -4g = -12 \quad \Rightarrow \quad g = 3 \]

### Step 6: Write the equation of the required circle.
Now substituting \( g = 3 \) and \( f = -\frac{3}{2} \) into the circle equation:
\[ x^2 + y^2 + 2(3)x + 2\left(-\frac{3}{2}\right)y = 0 \]
\[ x^2 + y^2 + 6x - 3y = 0 \]

### Step 7: Multiply through by 4 to eliminate fractions (if necessary).
To express it in a more standard form:
\[ 4x^2 + 4y^2 + 24x - 12y = 0 \]

### Final Answer:
The required equation of the circle is:
\[ 4x^2 + 4y^2 + 24x - 12y = 0 \]