AP EAMCET2011MathematicsCircle
If the circle x^2+y^2+8 x-4 y+c=0 touches the circle x^2+y^2+2 x+4 y-11=0 externally and cuts the circle x^2+y^2-6 x+8 y+k=0 orthogonally, then k is equal to
Options
- A59
- B-59
- C19
- D-19
Correct answer
B. -59
Step-by-step solution
Given that circle x^2+y^2+8 x-4 y+c=0 touch the circle x^2+y^2+2 x+4 y-11=0 where gathered C₁=(-4,2) r₁= 16+4-c = 20-c C₂=(-1,-2) gathered and r₂= 1+4+11 =4 From Eq. (i), aligned & (-4+1)^2+(2+2)^2 & = 20-c +4 & 5= 20-c +4 & c=19 & aligned Also, the circle x^2+y^2+8 x-4 y+c=0 cuts the circles x^2+y^2-6 x+8 y+k=0 orthogonally, then c₁ (-4,2) aligned & C₃ (3,-4) & (C₁ C₃ )^2= (r₁ )^2+ (r₃ )^2 & where, & r₁= 16+4-c & r₃= 9+16-k & & (-4-3)^2+(2+4)^2 =(20-c)+(25-k) & 49+36=45-k-c & k+c=-40 & k+19=-40 & k=-59 & aligned