The common tangents to the circles `x^(2) +y^(2) +2x = 0` and `x^(2) +y^(2) -6x= 0` from a triangle which is `:`

Options

A. equilateral
B. isosceles
C. right angled
D. None of these

Answer

To find the common tangents to the circles given by the equations \(x^2 + y^2 + 2x = 0\) and \(x^2 + y^2 - 6x = 0\), we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
1. **First Circle**:
\[ x^2 + y^2 + 2x = 0 \implies (x^2 + 2x + 1) + y^2 = 1 \implies (x + 1)^2 + y^2 = 1^2 \]
This circle has center \((-1, 0)\) and radius \(1\).

2. **Second Circle**:
\[ x^2 + y^2 - 6x = 0 \implies (x^2 - 6x + 9) + y^2 = 9 \implies (x - 3)^2 + y^2 = 3^2 \]
This circle has center \((3, 0)\) and radius \(3\).

### Step 2: Identify the centers and radii
- Center of the first circle \(C_1 = (-1, 0)\), radius \(r_1 = 1\).
- Center of the second circle \(C_2 = (3, 0)\), radius \(r_2 = 3\).

### Step 3: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1\) and \(C_2\) is given by:
\[ d = \sqrt{(3 - (-1))^2 + (0 - 0)^2} = \sqrt{(3 + 1)^2} = \sqrt{4^2} = 4 \]

### Step 4: Determine the type of triangle formed by the common tangents
The lengths of the tangents from the centers to the points of tangency can be calculated using the formula for the length of the tangent from a point to a circle:
\[ L = \sqrt{d^2 - (r_1 + r_2)^2} \]
Since we are looking for common tangents, we will consider the external tangents:
- The distance between the centers \(d = 4\)
- The sum of the radii \(r_1 + r_2 = 1 + 3 = 4\)

### Step 5: Analyze the triangle formed by the tangents
Since \(d = r_1 + r_2\), the common tangents form a triangle with the centers of the circles. The triangle formed by the points of tangency and the centers of the circles will be a right triangle.

### Step 6: Calculate angles of the triangle
Using the properties of the triangle:
- The angle at the center corresponding to the tangent can be calculated. Since the tangents touch the circles at right angles, we can conclude that the angles at the points of tangency are \(90^\circ\).

### Conclusion
The triangle formed by the common tangents to the two circles is a right triangle.