Consider the given equation. x2+y2−6x+4y=12 (x−3)2+(y+2)2=25 (x−3)2+(y+2)2=(5)2 If m is the slope, then the equation of the tangent is given by, y+2=m(x−3)±5√m2+1 The slope of the tangent is same as the line 4x+3y+5=0 Thus, m=−
4
3
y+2=−
4
3
(x−3)±5√(−
4
3
)2+1 y+2=
−4
3
(x−3)±5(
5
3
) 3y+6=−4x+12±25 4x+3y=6±25 Therefore, the equation becomes 4x+3y−31=0 4x+3y+19=0