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过点A(0,6)且与圆C:x2+y2+10x+10y=0切于原点的圆的方程为.

题目详情
过点A(0,6)且与圆C:x 2 +y 2 +10x+10y=0切于原点的圆的方程为______.
▼优质解答
答案和解析
由圆C:x 2 +y 2 +10x+10y=0化为(x+5) 2 +(y+5) 2 =50.可得圆心C(-5,-5),半径R=5
2

设所求的圆的标准方程为(x-a) 2 +(y-b) 2 =r 2 ,∵此圆过点A(0,6)且与圆C:x 2 +y 2 +10x+10y=0切于原点,
a 2 +(6-b ) 2 = r 2
a 2 + b 2 = r 2
(a+5 ) 2 +(b+5 ) 2 =(5
2
+r ) 2
解得
a=3
b=3
r 2 =18

故所求的圆的方程为(x-3) 2 +(y-3) 2 =18.
故答案为(x-3) 2 +(y-3) 2 =18.