question_answer

Tangents drawn from point of intersection A of circles \[{{x}^{2}}+{{y}^{2}}=4\]and \[{{(x-\sqrt{3})}^{2}}+\]\[{{(y-3)}^{2}}=4\] cut the line joining their centres at B and C. Triangle BAG is

A)  equilateral triangle

B)  right angled triangle

C)  obtuse angled triangle

D)  isosceles triangle and\[\angle ABC=\frac{\pi }{6}\]

Correct Answer: A

Solution :

 Radius of the circles are same. \[\Rightarrow \]               \[AB=AC\] Also if \[\theta \]is the angle between the tangents then \[\cos \theta =\frac{12-4-4}{2(2)(2)}=\frac{1}{2}\] \[\Rightarrow \]               \[\theta =\frac{\pi }{3}\] Hence, \[\Delta ABC\]is an equilateral triangle.