直線及圓形考試問題

(a)
C: x² + y² - 4x - 6y + 12 = 0
x² - 4x + 4 + y² - 6y + 9 = 1
(x - 2)² + (y - 3)² = 1²

Centre = (2, 3)
Radius = 1

(b)
Let D be the center of C.

OP² = a² + b²
PQ² = PD² - QD²
　　= (a - 2)² + (b - 3)² - 1²
　　= a² - 4a + 4 + b² - 6b + 9 - 1
　　= a² + b² - 4a - 6b + 12

OP = PQ means OP² = PQ²
a² + b² = a² + b² - 4a - 6b + 12
- 4a - 6b + 12 = 0
2a + 3b - 6 = 0
2a + 3b = 6

(c)
P(a,b) is on the y-axis, then a = 0, b = 2.
PQ² = a² + b² - 4a - 6b + 12
　　= 0 + 4 - 0 - 12 + 12
　　= 4
PQ = 2

(d)
PQ attains minimum when PQ² attains minimum.
Consider PQ²
= a² + b² - 4a - 6b + 12
= a² + [(6-2a)/3]² - 4a - 6[(6-2a)/3] + 12
= a² + (6-2a)²/9 - 4a - 2(6-2a) + 12
= (1/9) { 9a² + (6-2a)² - 36a - 18(6-2a) + 108 }
= (1/9) { 9a² + 36 - 24a + 4a² - 36a - 108 + 36a + 108 }
= (1/9) { 13a² - 24a + 36 }
= (13/9) { a² - (24/13)a + 36/13 }
= (13/9) { a² - 2(12/13)a + (12/13)² - (12/13)² + 36/13 }
= (13/9) { (a - 12/13)² + 324/169 }
= (13/9)(a - 12/13)² + 36/13

PQ² attains minimum when a = 12/13,
b = (6-2a)/3 = 18/13.

Therefore, when PQ attains minimum, P = (12/13, 18/13)