Let the point P be the vertex of the parabola y=x2-6x+12. If a line passing through the point P intersects the circle x2+y2-2x-4y+3=0 at the points R and S, then the maximum value of (PR+PS)2 is :
- 10
- 20
- 25
- 5
Answer
(2) 20
Solution
[figure]
y=x2-6x+12=(x-3)2+3 vertex P(3, 3)
Circle: (x-1)2+(y-2)2=2 centre C(1, 2), radius 2
PC= 4+1= 5> 2, P lies outside the circle, so R and S are on the same side of P
Let M be the mid-point of chord RS (foot of the perpendicular from C), MR=MS:
PR+PS=(PM-MR)+(PM+MS)=2 PM
PM2=PC2-CM2 PC2=5 (equality when the line passes through C)
(PR+PS)2=4 PM2 20
Maximum value =20
Answer: (2)