Let | z-i2 z+i |= 13, z C, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0,0), C and ( ,0) is 11 square units, then 2 equals:
- 12125
- 100
- 8125
- 50
Answer
(2) 100
Solution
3| z-i|=|2 z+i| 3|z+i|=|2z-i| (a conjugate has the same modulus)
9 [x2+(y+1)2 ]=4x2+(2y-1)2 x2+y2+ 225y+ 85=0, centre C (0,- 115 )
[figure]
Right angle at O: 12 115 | |=11 | |=10
2=100
Answer: (2)