Persamaan garis singgung lingkaran (x+4)2+(y−1)2=8\left(x+4\right)^2+\left(y-1\right)^2=8(x+4)2+(y−1)2=8 di titik (−2,3)\left(-2,3\right)(−2,3) adalah ....
x−y−1=0x-y-1=0x−y−1=0
x+y+1=0x+y+1=0x+y+1=0
x+y−1=0x+y-1=0x+y−1=0
2x−y+1=02x-y+1=02x−y+1=0
2x+y−1=02x+y-1=02x+y−1=0