Question

求过点M(-2,1)且与圆c:(x+1)²+(y+2)²=1相切的直线方程?

Answer 1

Ax＋By＋C=0,即－2A＋B＋C＝0那么直线系方程X2＋Y2＋2X＋4Y＋4＝0,同解,则：X2＋Y2＋2X＋4Y＋4十入（AX＋By＋C)＝X2＋Y2＋(2＋入A）x＋(4＋入B)＋(C＋入)=0,所以1,2＋入A＝A,4＋入B=B,C＋入＝C,－2A＋B＋C＝0,A＝2,B＝4,入＝0,C＝0,2x＋4y＝0