Question

已知点A(-3,4),求过点A点与圆x平方+y平方-2x+4y-11=0相切的直线方程

Answer 1

y-4=k(x+3)与圆有一个交点

y=kx+3k+4 x^2+(kx+3k+4)^2-2x+4(kx+3k+4)-11=0

(k^2+1)x^2+[2k(3k+4)+4k]x+(3k+4)^2+4(3k+4)-11=0

(k^2+1)x^2+(6k^2+12k-2)x+(9k^2+36k+23)=0

判别式=(6k^2+12k-2)^2-4(k^2+1)(9k^2+36k+23)=0

k^2+24k+11=0 求出K后代 入y=kx+3k+4即可