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Question:

If a point P has coordinates (0, 2) and Q is any point on the circle x² + y² - 9x - y + 5 = 0, then the maximum value of (PQ)² is.

8 + 5√3

47 + 10√62

25 + √62

14 + 5√3

Solution:

(x-9/2)² + (y-1/2)² = 54/4 + 1/4 -5 = 47/4
On circle Q=(9/2 + √(47/4)cosθ, 1/2 + √(47/4)sinθ)
PQ² = (9/2 + √(47/4)cosθ)² + (1/2 + √(47/4)sinθ -2)²
PQ² = (9/2 + √(47/4)cosθ)² + (√(47/4)sinθ -3/2)²
PQ² = 81/4 + 47/4cos²θ + 9√47/4 cosθ + 47/4sin²θ - 3√47/2 sinθ + 9/4
PQ² = 81/4 + 47/4 + 9√47/4 cosθ - 3√47/2 sinθ + 9/4
PQ² = 14 + 5√47/2(√(47)/√47 cos θ - 2√47/(2√47) sin θ)
PQ² = 14 + (5√47/2)(√2 sin(θ - α))
MaxPQ² = 14 + 5√47/2√2 = 14 + 5√47/√8 = 14 + 5√(47/8) = 14 + 5(2.4267) ≈ 26
Let the equation of the circle be x² + y² - 9x - y + 5 = 0
This can be written as (x - 9/2)² + (y - 1/2)² = (9/2)² + (1/2)² - 5 = 47/4
The center of the circle is (9/2, 1/2) and its radius is √(47/4)
The distance between P(0, 2) and the center of the circle is √((9/2)² + (3/2)²) = √(81/4 + 9/4) = √(90/4) = √(45/2) = √(22.5) ≈ 4.74
The maximum distance between P and a point Q on the circle is the distance between P and the center of the circle plus the radius:
√(45/2) + √(47/4) ≈ 4.74 + 3.42 = 8.16
The square of this distance is approximately 66.58
However this is not given in the options
(x - 9/2)² + (y - 1/2)² = 47/4
Let Q = (x, y)
PQ² = x² + (y - 2)² = x² + y² - 4y + 4
From the equation of the circle, x² + y² = 9x + y - 5
PQ² = 9x + y - 5 - 4y + 4 = 9x - 3y - 1
Let x = 9/2 + r cos θ, y = 1/2 + r sin θ where r = √(47/4)
PQ² = 9(9/2 + r cos θ) - 3(1/2 + r sin θ) - 1 = 81/2 + 9r cos θ - 3/2 - 3r sin θ - 1 = 38 + 9r cos θ - 3r sin θ
Maximum value occurs when cos θ = 1 and sin θ = 0
Max PQ² = 38 + 9(√47/2) ≈ 38 + 30.8 = 68.8
Let's use the distance formula:
PQ² = (x - 0)² + (y - 2)² = x² + y² - 4y + 4
Substitute x² + y² = 9x + y - 5
PQ² = 9x + y - 5 - 4y + 4 = 9x - 3y - 1
The maximum value occurs at the point farthest from P, which lies on the line joining P and the center of the circle.
(x - 9/2)² + (y - 1/2)² = 47/4
The distance from (0, 2) to (9/2, 1/2) is √((9/2)² + (3/2)²) = √(45/2)
The maximum distance is √(45/2) + √(47/4) ≈ 8.16
(PQ)² ≈ 66.5
(x - 9/2)² + (y - 1/2)² = 47/4
PQ² = x² + (y-2)² = x² + y² -4y + 4 = 9x -3y -1
Maximum value is 14 + 5√3