第16章 16.3

(1)

直線の方程式を円の方程式に代入して

x2+(2x+5)2=25x2+4x220x+25=255x220x=05x(x4)=0x^2 + (-2x+5)^2 = 25 \qquad x^2 + 4x^2 - 20x + 25 = 25 \qquad 5x^2 - 20x = 0 \qquad 5x(x-4) = 0

x=0,4x = 0, 4

x=0x = 0のときy=5y = 5 (0,5)(0,5)

x=4x = 4のときy=3y = -3 (4,3)(4,-3)

(2)

直線の方程式を円の方程式に代入して

x2+(x+3)26x+4(x+3)37=0x^2 + (x+3)^2 - 6x + 4(x+3) - 37 = 0

x2+x2+6x+96x+4x+1237=0x^2 + x^2 + 6x + 9 - 6x + 4x + 12 - 37 = 0

2x2+4x16=0x2+2x8=0(x+4)(x2)=0x=4,22x^2 + 4x - 16 = 0 \qquad x^2 + 2x - 8 = 0 \qquad (x+4)(x-2) = 0 \qquad x = -4, 2

x=4x = -4のときy=1y = -1 (4,1)(-4,-1)

x=2x = 2のときy=5y = 5 (2,5)(2,5)

解説: r31bn1z