To solve the problem step by step, we will follow these steps:
### Step 1: Write the equations of the circles
The given equations of the circles are:
1. \( x^2 + y^2 - 2x - 4y - 20 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 6x + 2y - 90 = 0 \) (Circle 2)
### Step 2: Convert the equations to standard form
To convert the equations into standard form, we will complete the square.
**For Circle 1:**
\[
x^2 - 2x + y^2 - 4y = 20
\]
Completing the square:
\[
(x - 1)^2 - 1 + (y - 2)^2 - 4 = 20
\]
\[
(x - 1)^2 + (y - 2)^2 = 25
\]
Thus, the center \(O_1\) is \((1, 2)\) and the radius \(R_1 = 5\).
**For Circle 2:**
\[
x^2 + 6x + y^2 + 2y = 90
\]
Completing the square:
\[
(x + 3)^2 - 9 + (y + 1)^2 - 1 = 90
\]
\[
(x + 3)^2 + (y + 1)^2 = 100
\]
Thus, the center \(O_2\) is \((-3, -1)\) and the radius \(R_2 = 10\).
### Step 3: Calculate the distance between the centers
The distance \(d\) between the centers \(O_1(1, 2)\) and \(O_2(-3, -1)\) is calculated as follows:
\[
d = \sqrt{(1 - (-3))^2 + (2 - (-1))^2} = \sqrt{(1 + 3)^2 + (2 + 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]
### Step 4: Check if the circles touch each other
The circles touch each other if the distance between their centers equals the difference of their radii:
\[
d = R_2 - R_1 \Rightarrow 5 = 10 - 5
\]
Since this condition holds, the circles touch each other internally.
### Step 5: Find the point of contact
Let the point of contact be \(A(h, k)\). Since \(O_1\) is the midpoint of \(O_2\) and \(A\), we can write:
\[
(1, 2) = \left(\frac{h - 3}{2}, \frac{k - 1}{2}\right)
\]
This gives us two equations:
1. \(1 = \frac{h - 3}{2} \Rightarrow h - 3 = 2 \Rightarrow h = 5\)
2. \(2 = \frac{k - 1}{2} \Rightarrow k - 1 = 4 \Rightarrow k = 5\)
Thus, the coordinates of the point of contact are \(A(5, 5)\).
### Step 6: Find the equation of the common tangent
To find the equation of the tangent at the point of contact \(A(5, 5)\) for Circle 1, we first find the slope of the tangent line.
Differentiating the equation of Circle 1:
\[
2x + 2y \frac{dy}{dx} - 4 \frac{dy}{dx} = 0
\]
\[
\Rightarrow (2y - 4) \frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = \frac{-2x}{2y - 4}
\]
Substituting \(x = 5\) and \(y = 5\):
\[
\frac{dy}{dx} = \frac{-2(5)}{2(5) - 4} = \frac{-10}{10 - 4} = \frac{-10}{6} = -\frac{5}{3}
\]
The slope of the tangent at point \(A(5, 5)\) is \(-\frac{5}{3}\).
Using the point-slope form of the equation of a line:
\[
y - 5 = -\frac{5}{3}(x - 5)
\]
Multiplying through by 3 to eliminate the fraction:
\[
3(y - 5) = -5(x - 5)
\]
\[
3y - 15 = -5x + 25
\]
Rearranging gives:
\[
5x + 3y - 40 = 0
\]
### Final Result
The coordinates of the point of contact are \((5, 5)\) and the equation of the common tangent is:
\[
5x + 3y - 40 = 0
\]
