Let P be a moving point on the circle x2+y2-6x-8y+21=0. Then, the maximum distance of P from the vertex of the parabola x2+6x+y+13=0 is equal to:
- 8
- 10
- 12
- 9
Answer
(3) 12
Solution
[figure]
Circle: (x-3)2+(y-4)2=4 centre C(3, 4), radius r=2
Parabola: x2+6x+9=-y-4 (x+3)2=-(y+4)
vertex V(-3, -4)
VC= (3+3)2+(4+4)2= 36+64=10
V lies outside the circle, the farthest point P is on line VC beyond C
VP =VC+r=10+2=12
Answer: (3)