AP EAMCET202420 May 2024Morning ShiftMathematicsCircleActual
If the circles x^2+y^2-8 x-8 y+28=0 and x^2+y^2-8 x-6 y+25- ^2=0 have only one common tangent, then =
Options
- A=4
- B=2
- C=1
- D=5
Correct answer
C. =1
Step-by-step solution
S₁=x^2+y^2-8 x-8 y+28=0 C₁=(4,4), r₁= 16+16-28 =2S₂=x^2+y^2-8 x-6 y+25- ^2=0 C₂=(4,3), r₂= 16+9-25+ ^2 = Condition for one common tangent is |r₁-r₂ |=C₁ C₂ (2- )^2=1 =1