AP EAMCET201922 Apr 2019Morning ShiftMathematicsCircleActual
The point of intersection of the common tangents drawn to the circles (x^2+y^2-4 x-2 y+1=0 ) and (x^2+y^2-6 x-4 y+4=0 ), is
Options
- A( ( 5 2 , 3 2 ) )
- B( ( 6 5 , 1 5 ) )
- C((0,-1) )
- D( ( 12 5 , 7 5 ) )
Correct answer
C. ((0,-1) )
Step-by-step solution
Given equation of circles are ( aligned & x^2+y^2-4 x-2 y+1=0 (i) & and x^2+y^2-6 x-4 y+4=0 (ii) aligned ) Here, (x^2+y^2-6 x-4 y+4=0 ) ( aligned & C₁=(2,1), C₂=(3,2) & r₁= 4+1-1 = 4 =2 & and r₂= 9+4-4 = 9 =3 aligned ) and (r₂= 9+4-4 = 9 =3 ) Now, ( aligned C₁ C₂ & = (3-2)^2+(2-1)^2 & = 1+1 = 2 aligned ) ( array ll and & r₁+r₂=2+3=5 & C₁ C₂ Hence, (P(x, y)=(0,-1) )