Find the equation of the circle passing through the points of contact of the direct common tangent of `x^2+y^2=16` and `x^2+y^2-12x+32=0`

Answer

To find the equation of the circle passing through the points of contact of the direct common tangent of the circles given by \(x^2 + y^2 = 16\) and \(x^2 + y^2 - 12x + 32 = 0\), we can follow these steps:

### Step 1: Identify the centers and radii of the circles
1. The first circle \(x^2 + y^2 = 16\) has its center at \(O(0, 0)\) and radius \(r_1 = 4\) (since \(\sqrt{16} = 4\)).
2. The second circle can be rewritten as \(x^2 + y^2 - 12x + 32 = 0\). Completing the square for \(x\):
\[ (x - 6)^2 + y^2 = 4 \]
This gives us the center \(A(6, 0)\) and radius \(r_2 = 2\) (since \(\sqrt{4} = 2\)).

### Step 2: Find the external center of the circles
The external center \(S_2\) divides the line segment joining \(O(0, 0)\) and \(A(6, 0)\) in the ratio of their radii, which is \(r_1 : r_2 = 4 : 2 = 2 : 1\).
Using the section formula:
\[ S_2 = \left( \frac{2 \cdot 6 + 1 \cdot 0}{2 + 1}, \frac{2 \cdot 0 + 1 \cdot 0}{2 + 1} \right) = \left( \frac{12}{3}, 0 \right) = (4, 0) \]

### Step 3: Find the equation of the direct common tangent
The equation of the direct common tangent can be derived using the formula for the tangent to two circles:
\[ y = mx + c \]
For the circles, we can find the slope \(m\) and the intercept \(c\). The equation of the tangent can be expressed as:
\[ L: 3x - 4 = 0 \]

### Step 4: Find the chord of contact for the second circle
The chord of contact from point \(S_2(4, 0)\) with respect to the second circle \(x^2 + y^2 - 12x + 32 = 0\) is given by:
\[ x^2 + y^2 - 12x + 32 + \lambda(3x - 4) = 0 \]
This simplifies to:
\[ x^2 + y^2 + (3\lambda - 12)x + (32 - 4\lambda) = 0 \]

### Step 5: Set up the equations for the circle
We now have two equations representing the same circle:
1. From the first circle:
\[ x^2 + y^2 - 16 + \mu(3x - 20) = 0 \]
2. From the second circle:
\[ x^2 + y^2 - 12x + 32 + \lambda(3x - 4) = 0 \]

### Step 6: Equate coefficients
Since both equations represent the same circle, we equate coefficients:
1. Coefficient of \(x\):
\[ 3\lambda = 3\mu - 12 \]
2. Constant term:
\[ -16 + 32 - 4\lambda = 0 \implies 16 - 4\lambda = 0 \implies \lambda = 4 \]

Substituting \(\lambda = 4\) into the first equation gives:
\[ 3(4) = 3\mu - 12 \implies 12 = 3\mu - 12 \implies 3\mu = 24 \implies \mu = 8 \]

### Step 7: Final equation of the circle
Substituting \(\lambda\) and \(\mu\) back into either equation gives:
\[ x^2 + y^2 - 16 - 2(3x - 4) = 0 \implies x^2 + y^2 - 6x - 8 = 0 \]

### Conclusion
Thus, the equation of the circle passing through the points of contact of the direct common tangent is:
\[ x^2 + y^2 - 6x - 8 = 0 \]