Circle

2x - y =1 <=>y=2x-1=>slope=2
let the slope of the tangent oto the circle is y=2x+C

x^2 + y^2 + 4x - 6y + 5 = 0
x^2+(2x+C)^2+4x-6(2x+C)+5=0
x^2+4x^2+4Cx+C^2+4x-12x-6C+5=0
5x^2+(4C-8)x+(C^2-6C+5)=0
D=0
(4C-8)^2-4(5)(C^2-6C+5)=0
16C^2-64C+64-20C^2+120C-100=0
-4C^2+56C-36=0
C^2-14+9=0
C=13.3 or 0.68
So the slope of the circleis y=2x+13.3 anf y=2x+0.68//

設2x-y+c=0
圓心與tangent的距離=半徑
∴∣[2(-2)-3+c] / √(2^2+1^2) ∣=√[(-2)^2+3^2-5]
(c-7)^2 / 5=8 (two side ^2)
(c-7)=±√40
c=±√40+7

∴equation of the tangent
2x- y ± √40+7=0