The point of intersection of common transverse tangents of two circles `x^2+ y^2 - 24x +2y +120= 0` and `x^2 + y^2 +20 x - 6y- 116 = 0` is

Answer

To find the point of intersection of the common transverse tangents of the two circles given by the equations:

1. \( x^2 + y^2 - 24x + 2y + 120 = 0 \)
2. \( x^2 + y^2 + 20x - 6y - 116 = 0 \)

we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form

**For Circle 1:**
The first circle can be rewritten by completing the square.

Starting with:
\[ x^2 - 24x + y^2 + 2y + 120 = 0 \]

Completing the square for \(x\):
\[ x^2 - 24x = (x - 12)^2 - 144 \]

Completing the square for \(y\):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting back:
\[ (x - 12)^2 - 144 + (y + 1)^2 - 1 + 120 = 0 \]
\[ (x - 12)^2 + (y + 1)^2 - 25 = 0 \]
Thus, the first circle is:
\[ (x - 12)^2 + (y + 1)^2 = 25 \]
This gives us the center \(C_1(12, -1)\) and radius \(r_1 = 5\).

**For Circle 2:**
The second circle can be rewritten similarly.

Starting with:
\[ x^2 + 20x + y^2 - 6y - 116 = 0 \]

Completing the square for \(x\):
\[ x^2 + 20x = (x + 10)^2 - 100 \]

Completing the square for \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting back:
\[ (x + 10)^2 - 100 + (y - 3)^2 - 9 - 116 = 0 \]
\[ (x + 10)^2 + (y - 3)^2 - 225 = 0 \]
Thus, the second circle is:
\[ (x + 10)^2 + (y - 3)^2 = 225 \]
This gives us the center \(C_2(-10, 3)\) and radius \(r_2 = 15\).

### Step 2: Use the section formula to find the point of intersection

The point \(P\) of intersection of the common transverse tangents divides the line segment \(C_1C_2\) in the ratio of the radii of the circles, which is \(r_1 : r_2 = 5 : 15 = 1 : 3\).

Using the section formula:
\[ P_x = \frac{m x_2 + n x_1}{m+n} \quad \text{and} \quad P_y = \frac{m y_2 + n y_1}{m+n} \]
where \(m = 1\), \(n = 3\), \(C_1(12, -1)\) and \(C_2(-10, 3)\).

Calculating \(P_x\):
\[ P_x = \frac{1 \cdot (-10) + 3 \cdot 12}{1 + 3} = \frac{-10 + 36}{4} = \frac{26}{4} = \frac{13}{2} \]

Calculating \(P_y\):
\[ P_y = \frac{1 \cdot 3 + 3 \cdot (-1)}{1 + 3} = \frac{3 - 3}{4} = \frac{0}{4} = 0 \]

### Final Result:
Thus, the point of intersection of the common transverse tangents is:
\[ P\left(\frac{13}{2}, 0\right) \]