Tangents are drawn from a point on the circle `x^(2)+y^(2)-4x+6y-37=0` to the circle `x^(2)+y^(2)-4x+6y-12=0`. The angle between the tangents, is

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Explanation

To solve the problem of finding the angle between the tangents drawn from a point on one circle to another circle, we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the circles given are:
1. Circle 1: \( x^2 + y^2 - 4x + 6y - 37 = 0 \)
2. Circle 2: \( x^2 + y^2 - 4x + 6y - 12 = 0 \)

### Step 2: Rewrite the equations in standard form
We will complete the square for both circles.

**For Circle 1:**
\[ x^2 - 4x + y^2 + 6y - 37 = 0 \]
Completing the square:
- For \(x^2 - 4x\):
\[ = (x - 2)^2 - 4 \]
- For \(y^2 + 6y\):
\[ = (y + 3)^2 - 9 \]

Substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 37 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 50 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 50 \]
This represents a circle with center \((2, -3)\) and radius \(5\sqrt{2}\).

**For Circle 2:**
\[ x^2 - 4x + y^2 + 6y - 12 = 0 \]
Completing the square:
- For \(x^2 - 4x\):
\[ = (x - 2)^2 - 4 \]
- For \(y^2 + 6y\):
\[ = (y + 3)^2 - 9 \]

Substituting back into the equation:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 25 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 25 \]
This represents a circle with center \((2, -3)\) and radius \(5\).

### Step 3: Identify the relationship between the circles
Notice that the center of both circles is the same, \((2, -3)\). The radius of Circle 1 is \(5\sqrt{2}\) and the radius of Circle 2 is \(5\).

### Step 4: Determine the angle between the tangents
Since the tangents are drawn from a point on Circle 1 (which is the same as the center of both circles) to Circle 2, and since Circle 1 is the director circle of Circle 2, the angle between the tangents will be \(90^\circ\).

### Final Answer
The angle between the tangents is \(90^\circ\).

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