If P(6, 1) be the orthocentre of the triangle whose vertices are A(5, -2), B(8, 3) and C(h, k), then the point C lies on the circle :
- x2 + y2 - 74 = 0
- x2 + y2 - 65 = 0
- x2 + y2 - 61 = 0
- x2 + y2 - 52 = 0
Answer
(2) x2 + y2 - 65 = 0
Solution
[figure]
∵ the orthocentre P lies on every altitude:
mAP = 1 + 26 - 5 = 3, BC ⊥ AP: y - 3 = - 13(x - 8) x + 3y - 17 = 0
mBP = 1 - 36 - 8 = 1, AC ⊥ BP: y + 2 = -(x - 5) x + y - 3 = 0
Solving: 2y = 14 y = 7, x = -4 ⇒ C(-4, 7)
h2 + k2 = 16 + 49 = 65 ⇒ C lies on x2 + y2 - 65 = 0
Answer: (2)