The common tangent at the point of contact of the two circles `x^2+y^2-2x-4y-20=0, x^2+y^2+6x+2y-90=0` is

Options

A. 4x+3y+35=0
B. 3x+4y+35=0
C. 4x+3y-35=0
D. 4x-2y-110 =0

Answer

To find the common tangent at the point of contact of the two circles given by the equations \(x^2 + y^2 - 2x - 4y - 20 = 0\) and \(x^2 + y^2 + 6x + 2y - 90 = 0\), we can follow these steps:

### Step 1: Rewrite the equations of the circles
We start by rewriting the equations of the circles in standard form.

1. For the first circle:
\[ x^2 + y^2 - 2x - 4y - 20 = 0 \]
Completing the square:
\[ (x^2 - 2x) + (y^2 - 4y) = 20 \]
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 20 \]
\[ (x - 1)^2 + (y - 2)^2 = 25 \]
This represents a circle with center \((1, 2)\) and radius \(5\).

2. For the second circle:
\[ x^2 + y^2 + 6x + 2y - 90 = 0 \]
Completing the square:
\[ (x^2 + 6x) + (y^2 + 2y) = 90 \]
\[ (x + 3)^2 - 9 + (y + 1)^2 - 1 = 90 \]
\[ (x + 3)^2 + (y + 1)^2 = 100 \]
This represents a circle with center \((-3, -1)\) and radius \(10\).

### Step 2: Find the equation of the common tangent
The equation of the common tangent to two circles can be found using the formula:
\[ s_1 - s_2 = 0 \]
where \(s_1\) and \(s_2\) are the equations of the circles.

Substituting the equations:
\[ s_1 = x^2 + y^2 - 2x - 4y - 20 \]
\[ s_2 = x^2 + y^2 + 6x + 2y - 90 \]

Now, we set up the equation:
\[ s_1 - s_2 = 0 \]
This gives:
\[ (x^2 + y^2 - 2x - 4y - 20) - (x^2 + y^2 + 6x + 2y - 90) = 0 \]

### Step 3: Simplify the equation
Simplifying the above equation:
\[ -2x - 4y - 20 - 6x - 2y + 90 = 0 \]
Combining like terms:
\[ -8x - 6y + 70 = 0 \]

### Step 4: Rearranging the equation
Now, we can rearrange the equation:
\[ 8x + 6y = 70 \]
Dividing the entire equation by 2:
\[ 4x + 3y = 35 \]

### Step 5: Final form of the equation
Rearranging gives us:
\[ 4x + 3y - 35 = 0 \]

### Conclusion
Thus, the equation of the common tangent at the point of contact of the two circles is:
\[ \boxed{4x + 3y - 35 = 0} \]