Locus of point of intersection of perpendicular tangents to the circle `x^(2)+y^(2)-4x-6y-1=0` is

Options

A. `x^(2)+y^(2)-4x-6y-15=0`
B. `x^(2)+y^(2)-4x-6y+15=0`
C. `x^(2)+y^(2)-4x-3y-15=0`
D. `x^(2)+y^(2)+4x+6y-15=0`

Answer

To find the locus of the point of intersection of perpendicular tangents to the circle given by the equation \(x^2 + y^2 - 4x - 6y - 1 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the given equation of the circle in standard form. The equation is:
\[ x^2 + y^2 - 4x - 6y - 1 = 0 \]
We can rearrange this to group the \(x\) and \(y\) terms:
\[ x^2 - 4x + y^2 - 6y = 1 \]

### Step 2: Complete the Square
Next, we will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

For \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Now substituting these back into the equation:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 1 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 - 13 = 1 \]
Thus, we have:
\[ (x - 2)^2 + (y - 3)^2 = 14 \]
This shows that the center of the circle is \((2, 3)\) and the radius \(r = \sqrt{14}\).

### Step 3: Find the Director Circle
The locus of the intersection of perpendicular tangents to a circle is given by the equation of the director circle. The formula for the director circle is:
\[ (x - x_1)^2 + (y - y_1)^2 = 2r^2 \]
where \((x_1, y_1)\) is the center of the circle and \(r\) is the radius.

Substituting \(x_1 = 2\), \(y_1 = 3\), and \(r = \sqrt{14}\):
\[ (x - 2)^2 + (y - 3)^2 = 2(\sqrt{14})^2 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 = 28 \]

### Step 4: Expand the Equation
Now we will expand this equation:
\[ (x - 2)^2 + (y - 3)^2 = 28 \]
Expanding gives:
\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 28 \]
Combining like terms:
\[ x^2 + y^2 - 4x - 6y + 13 = 28 \]
Rearranging gives:
\[ x^2 + y^2 - 4x - 6y - 15 = 0 \]

### Final Result
Thus, the locus of the point of intersection of perpendicular tangents to the circle is:
\[ x^2 + y^2 - 4x - 6y - 15 = 0 \]