JEE Main2012MathematicsCircleActual

The number of common tangents of the circles given by x^2+y^2-8 x-2 y+1=0 and x^2+y^2+6 x+8 y=0 is

Options

Aone
Bfour
Ctwo
Dthree

Correct answer

C. two

Step-by-step solution

Given circles are x^2+y^2-8 x-2 y+1=0 and x^2+y^2+6 x+8 y=0 Their centres and radius are C₁(4,1), r₁= 16 =4 C₂(-3,-4), r₂= 25 =5 Now, C₁ C₂= 49+25 = 74 r₁-r₂=-1, r₁+r₂=9 Since, r₁-r₂ < C₁ C₂ < r₁+r₂ Number of common tangents =2

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