Let the line x-y=4 intersect the circle C: (x-4)2+(y+3)2=9 at the points Q and R. If P( , ) is a point on C such that PQ=PR, then (6 +8 )2 is equal to .
Answer
18
Solution
[figure]
Centre of C: (4, -3), radius 3
PQ=PR P lies on the perpendicular bisector of the chord QR, which passes through the centre and is perpendicular to x-y=4
its equation: x+y=1
Any point on it: (4+t, -3-t); on C t2+t2=9 t2= 92
P( , )=(4+t, -3-t)
6 +8 =6(4+t)+8(-3-t)=-2t
(6 +8 )2=4t2=4 92=18
Answer: 18