The number of common tangents to the circles $$x^{2} + y^{2} = 4$$​ and $$x^{2} + y^{2} − 8x + 12 = 0$$​ is:

Correct option is C

Given:
First circle: $$x^{2} + y^{2} = 4$$​​
Second circle: $$x^{2} + y^{2} − 8x + 12 = 0$$​​
Formula used:
If the distance between centers d and radii $$r_{1}, r_{2}$$​ satisfy:
$$d = r_{1} + r_{2}$$​=> circles touch externally => number of common tangents = 3
Solution:
For $$x^{2} + y^{2} = 4:$$​​
Center $$C_{1} = (0, 0),$$​ radius $$r_{1} = 2$$​​
For $$x^{2} + y^{2} − 8x + 12 = 0:$$​​
​$$(x − 4)^{2} + y^{2} = 4$$​​
Center $$C_{2} = (4, 0)$$​, radius $$r_{2} = 2$$​​
Distance between centers:
​$$d = \sqrt{[(4 − 0)^{2} + (0 − 0)^{2}]}\\d = 4$$​​
Since:
​$$d = r_{1} + r_{2} = 2 + 2 = 4$$​​
The circles touch each other externally.
Hence, the number of common tangents is 3.
The correct answer is (c) 3.