Tangents drawn from point of intersection A of circles `x^2+y^2=4 and (x-sqrt3)^2+(y-3)^2= 4` cut the line joinihg their centres at B and C Then triangle BAC is

Options

A. equilateral triangle
B. right angle triangle
C. obtuse angle triangle
D. isosceles triangle and `/_ABC = (pi)/(6)`

Explanation

To solve the problem, we need to analyze the given circles and the tangents drawn from their point of intersection. Let's break down the solution step by step.

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( x^2 + y^2 = 4 \) (Circle 1)
2. \( (x - \sqrt{3})^2 + (y - 3)^2 = 4 \) (Circle 2)

### Step 2: Find the centers and radii of the circles
- The center of Circle 1 is at \( (0, 0) \) with a radius of \( 2 \) (since \( r^2 = 4 \)).
- The center of Circle 2 is at \( (\sqrt{3}, 3) \) with a radius of \( 2 \) (since \( r^2 = 4 \)).

### Step 3: Find the points of intersection of the circles
To find the points of intersection, we can substitute \( y^2 = 4 - x^2 \) from Circle 1 into Circle 2's equation:
\[ (x - \sqrt{3})^2 + (4 - x^2 - 3)^2 = 4 \]
This simplifies to:
\[ (x - \sqrt{3})^2 + (1 - x^2)^2 = 4 \]
Expanding and solving this equation will yield the points of intersection.

### Step 4: Calculate the angle between the tangents
Let \( A \) be the point of intersection of the two circles. The tangents drawn from point \( A \) to both circles will form an angle \( \theta \) at point \( A \). The formula for the angle \( \theta \) between the tangents from point \( A \) to the circles can be derived using the cosine rule:
\[ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \]
where \( r_1 \) and \( r_2 \) are the radii of the circles, and \( d \) is the distance between the centers of the circles.

### Step 5: Calculate the distance between the centers
The distance \( d \) between the centers \( (0, 0) \) and \( (\sqrt{3}, 3) \) is:
\[ d = \sqrt{(\sqrt{3} - 0)^2 + (3 - 0)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \]

### Step 6: Substitute values into the cosine formula
Substituting \( r_1 = 2 \), \( r_2 = 2 \), and \( d = 2\sqrt{3} \) into the cosine formula gives:
\[ \cos \theta = \frac{2^2 + 2^2 - (2\sqrt{3})^2}{2 \cdot 2 \cdot 2} \]
\[ = \frac{4 + 4 - 12}{8} = \frac{-4}{8} = -\frac{1}{2} \]
This means \( \theta = \frac{2\pi}{3} \) (or \( 120^\circ \)).

### Step 7: Determine the type of triangle
Since the angle \( \theta \) between the tangents is \( 120^\circ \), the angles \( \angle BAC \) and \( \angle ABC \) can be determined. The triangle \( BAC \) is formed by the points \( A \), \( B \), and \( C \) where \( B \) and \( C \) are the points where the tangents intersect the line joining the centers.

Given that one angle is greater than \( 90^\circ \), triangle \( BAC \) is an obtuse triangle.

### Conclusion
The triangle \( BAC \) is an obtuse triangle.