一道数学题，请大家帮忙解一下，谢谢

Question

如题所述

Answer 1

解：

x
+
3y
-
5
=
0,x
+
3y
-
3
=
0
是平行线，

所以圆心在两条直线中间那条平行线上：x
+
3y
-
4
=
0
x
+
3y
=
4
2x
+
y
=
-1
联立得到圆心坐标：x
=
-7/5
，y
=
13/5
圆半径等于两条直线距离之半，注意到斜率
K
=
-1/3
tanA
=
-1/3
(sinA/cosA)^2
=
1/9
9*(sinA)^2
=
(cosA)^2
=
1
-
(sinA)^2
sinA
=
1/√10
R
=
(5
-
3)/2*sinA
=
1/√10
(x
+
7/5)^2
+
(y
-
13/5)^2
=
R^2

圆方程：
(x
+
7/5)^2
+
(y
-
13/5)^2
=
1/10