第16章 16.1

(1)

(x1)2+(y2)2=9(x-1)^2+(y-2)^2=9

(2)

r2=52+(1)2=26r^2=5^2+(-1)^2=26

(x5)2+(y+1)2=26(x-5)^2+(y+1)^2=26

(3)

r2=(1(3))2+(11)2=20r^2=\left(1-(-3)\right)^2+(-1-1)^2=20

(x1)2+(y+1)2=20(x-1)^2+(y+1)^2=20

(4)

中心は(1+(3)2,5+72)=(1,1)\left(\dfrac{1+(-3)}{2},\dfrac{-5+7}{2}\right)=(-1,1)

r2=(1(3))2+(57)24=40r^2=\dfrac{\left(1-(-3)\right)^2+(-5-7)^2}{4}=40

(x+1)2+(y1)2=40(x+1)^2+(y-1)^2=40

(5)

中心を(x,y)(x,y)とおくと

x2+y2=kx^2+y^2=k \qquad ①

x2+(y1)2=kx^2+(y-1)^2=k \qquad ②

(x2)2+(y2)2=k(x-2)^2+(y-2)^2=k \qquad ③

①=②から y=12y=\dfrac{1}{2}

①=③から x+y=2x+y=2 ④を代入して

x=32x=\dfrac{3}{2}

r2=(32)2+(12)2=104r^2 = \left(\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{10}{4}

(x32)2+(y12)2=104\left(x - \frac{3}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{10}{4}

解説: r31bn1z