已知M为圆C：X2+Y2-4X=0上任一点,且点Q（-2,3）.（1）/MQ/的最大值和最小值（2）若P在圆C上,求直线

圆C：x2 + y2 – 4x = 0 => (x – 2)2 + y2 = 4 ；
（1）点M在圆C上,三角代换设点M(2 + 2cosθ,2sinθ),θ∈[0,2π),所以|MQ| = √[(4 + 2cosθ)2 + (-3 + 2sinθ)2] => |MQ|2 = 16 + 16cosθ + 4cos2θ + 9 – 12sinθ + 4sin2θ = 29 + 16cosθ – 12sinθ = 29 + 20sin(θ + β),因为θ∈[0,2π) => sin(θ + β)∈[-1,1] => 20sin(θ + β)∈[-20,20] => 29 + 20sin(θ + β)∈[9,49],即|MQ|2∈[9,49] => |MQ|∈[3,7],所以|MQ|的最大值是7,最小值是3 ；
（2）点P在圆C上,三角代换设点P(x,y),则x2 + y2 – 4x = 0①,因为x∈[0,4],所以x≠-2,所以直线PQ的斜率存在,记k = kPQ = (y – 3)/(x + 2),所以y – 3 = k(x + 2) => y = kx + 2k + 3,代入①,可得x2 + (kx + 2k + 3)2 – 4x = 0 => (1 + k2)x2 + [2k(2k + 3) – 4]x + 4k2 + 12k + 9 = 0 => (1 + k2)x2 + 2(2k2 + 3k – 2)x + 4k2 + 12k + 9 = 0 => Δ= 4(2k2 + 3k – 2)2 – 4(1 + k2)(4k2 + 12k + 9) = 4(4k4 + 9k2 + 4 + 12k3 – 8k2 – 12k) – 4(4k4 + 12k3 + 9k2 + 4k2 + 12k + 9) = 4[(4k4 + 12k3 + k2 – 12k + 4) – (4k4 + 12k3 + 13k2 + 12k + 9)] = 4(-12k2 – 24k – 5) ≥ 0 => 12k2 + 24k + 5 ≤ 0②,令12k2 + 24k + 5 = 0,根的判别式△= 24*24 – 4*12*5 = 336,所以对应的一元二次方程的两根为k = (-24±4√21)/24 = (-6±√21)/6,所以不等式②的解集为k∈[(-6 - √21)/6,(-6 + √21)/6] .