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

Answer

To solve the problem of finding the type of triangle formed by 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
The equations of the circles can be rewritten in standard form by completing the square.

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

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

### Step 2: Identify the centers and radii
From the rewritten equations, we have:
- Circle 1 (C1): Center \((-1, 0)\), Radius \(1\)
- Circle 2 (C2): Center \((3, 0)\), Radius \(3\)

### Step 3: Determine the distance between the centers
The distance \(d\) between the centers of the two circles can be calculated using the distance formula:
\[ d = \sqrt{(3 - (-1))^2 + (0 - 0)^2} = \sqrt{(3 + 1)^2} = \sqrt{4^2} = 4 \]

### Step 4: Calculate the lengths of the tangents
The lengths of the common tangents can be calculated using the formula for the length of the tangent between two circles:
\[ L = \sqrt{d^2 - (r_1 + r_2)^2} \]
Where \(r_1\) and \(r_2\) are the radii of the circles.

Here, \(r_1 = 1\) and \(r_2 = 3\):
\[ L = \sqrt{4^2 - (1 + 3)^2} = \sqrt{16 - 16} = \sqrt{0} = 0 \]
This indicates that the circles are internally tangent to each other.

### Step 5: Analyze the triangle formed by the tangents
Since the common tangents are equal and the angles formed at the points of tangency are equal, the triangle formed by the tangents will be isosceles.

### Step 6: Find the angles of the triangle
Using the properties of the tangents and the geometry of the circles, we can find the angles:
1. The angle at the center of the first circle is \( \theta \).
2. The angle at the center of the second circle is also \( \theta \).
3. The angle between the tangents at the point of tangency is \( 180^\circ - 2\theta \).

### Step 7: Determine the type of triangle
Since the tangents are equal and the angles at the points of tangency are equal, the triangle formed is equilateral.

### Conclusion
The common tangents to the circles form an equilateral triangle.