Tentukan persamaan garis singgung lingkaran x² + y² + 2x - 8 = 0 dalam bentuk ax + by + c = 0 yang ditarik dari titik (4, 0) di luar lingkaran!

[tex]\displaystyle \text{misalkan garis singgungnya }y=mx +c\\\\y=mx+c\\0=4m+c\\-4m=c\\\\y=mx-4m\\\\x^2+y^2+2x-8=0\\x^2+(mx-4m)^2+2x-8=0\\x^2+m^2x^2-8m^2x+16m^2+2x-8=0\\(1+m^2)x^2+(2-8m^2)x+16m^2-8=0\\\\\text{agar menyinggung lingkaran, maka }\\D=0\\b^2-4ac=0\\(2-8m^2)^2-4(1+m^2)(16m^2-8)=0\\2^2(1-4m^2)^2-4(1+m^2)(16m^2-8)=0\\(1-4m^2)^2-(1+m^2)(16m^2-8)=0\\16m^4-8m^2+1=16m^4+8m^2-8\\-8m^2+1=8m^2-8\\0=16m^2-9\\0=(4m-3)(4m+3)\\m=\pm\frac34\\\\\therefore\\y=\frac34x-3\vee y=-\frac34x+3\\4y=3x-12\vee4y=-3x+12[/tex]
[tex]\displaystyle \boxed{\boxed{4y-3x+12=0\vee4y+3x-12=0}}[/tex]