JEE Math Practice Question
If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is :
- A.5
- B.2$$\sqrt {5} $$
- C.4
- D.$$\sqrt {37} $$
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