Question

過點A(-1,4),并且與圓(x-3)²+(y+1)²=5相切于點B(2,1)的圓的方程是___________.

Answer 1

(x-1)²+(y-3)²=5

設圓心為(a,b),
∵三點(3，-1)、(2,1)與(a,b)共線，
∴2a+b-5="0.                                                          " ①
又兩點A(-1,4)與B(2,1)都在圓上，
∴(a+1)²+(b-4)²=(a-2)²+(b-1)²,
即a-b+2="0.                                                            " ②
由①②聯立解得
半徑r=
∴所求圓的方程為(x-1)²+(y-3)²=5.