Given circle is x2+y2+6x+4y−3=0 Equation of the tangent at (x1,y1) is xx1+yy1+3(x+x1)+2(y+y1)−3=0 If this line passes through (1,−2), then x−2y+3(x+1)+2(y−2)−3=0 ⇒x−2y+3x+3+2y−4−3=0 ⇒4x−4=0 ⇒x−1=0 Equation of the normal is 0⋅x−1⋅y=λ It passes through (1,−2) −(−2)=λ⇒λ=2 ⇒−y=2⇒y+2=0