Locus of the points of intersection of perpendicular tangents drawn one to each of the circles `x^(2)+y^(2)-4x+6y-37=0, x^(2)+y^(2)-4x+6y-20=0` is

Options

A. `x^(2)+y^(2)-4x+6y=0`
B. `x^(2)+y^(2)-4x+6y-50=0`
C. `x^(2)+y^(2)-4x+6y-57=0`
D. `x^(2)+y^(2)-4x+6y-70=0`

Answer

To find the locus of the points of intersection of perpendicular tangents drawn to each of the circles given by the equations:

1. \( x^2 + y^2 - 4x + 6y - 37 = 0 \)
2. \( x^2 + y^2 - 4x + 6y - 20 = 0 \)

we will follow these steps:

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

**For the first circle:**
\[ x^2 + y^2 - 4x + 6y - 37 = 0 \]
We can complete the square for \(x\) and \(y\):
- For \(x\): \(x^2 - 4x = (x - 2)^2 - 4\)
- For \(y\): \(y^2 + 6y = (y + 3)^2 - 9\)

Substituting these into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 37 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 50 = 0 \]
Thus, the first circle is:
\[ (x - 2)^2 + (y + 3)^2 = 50 \]

**For the second circle:**
\[ x^2 + y^2 - 4x + 6y - 20 = 0 \]
Completing the square similarly:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 20 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 33 = 0 \]
Thus, the second circle is:
\[ (x - 2)^2 + (y + 3)^2 = 33 \]

### Step 2: Identify the centers and radii of the circles.

From the equations:
- The center of the first circle is \( (2, -3) \) with radius \( \sqrt{50} \).
- The center of the second circle is \( (2, -3) \) with radius \( \sqrt{33} \).

### Step 3: Determine the locus of the intersection of perpendicular tangents.

Let \( P(h, k) \) be the point of intersection of the perpendicular tangents to both circles. Since the tangents are perpendicular, we can use the property of rectangles formed by the tangents and the points of tangency.

### Step 4: Apply the Pythagorean theorem.

Let \( PA \) and \( PB \) be the lengths of the tangents from point \( P \) to the points of tangency on the circles. Since \( PA \) and \( PB \) are perpendicular, we can apply the Pythagorean theorem:
\[ PC^2 = PA^2 + AC^2 \]
Where \( AC \) is the distance between the points of tangency on the circles.

The distance \( AC \) can be calculated as:
\[ AC^2 = R_1^2 + R_2^2 = 50 + 33 = 83 \]
Thus, we have:
\[ PC^2 = PA^2 + 83 \]

### Step 5: Express \( PA^2 \) in terms of \( P \).

The distance \( PC \) can be expressed as:
\[ PC^2 = (h - 2)^2 + (k + 3)^2 \]

### Step 6: Set up the equation.

From the above, we have:
\[ (h - 2)^2 + (k + 3)^2 = PA^2 + 83 \]
Since \( PA^2 \) is the tangent squared from point \( P \) to the first circle, we can express it as:
\[ PA^2 = R_1^2 - (h - 2)^2 - (k + 3)^2 \]
Substituting this back into the equation gives:
\[ (h - 2)^2 + (k + 3)^2 = 50 + 83 \]

### Step 7: Final equation of the locus.

After simplification, we find:
\[ h^2 + k^2 - 4h + 6k - 70 = 0 \]
Thus, the locus of the points of intersection of the perpendicular tangents is:
\[ x^2 + y^2 - 4x + 6y - 70 = 0 \]