Given, circle is x2+y2+4x+6y−3=0 ...(i) Given, point is P(1,1). Since, equation of polar form of point (x1,y1) with respect to circle x2+y2+2gx+2fy+c=0 is
x.x1+y.y1+2g(
x+x1
2
)+2f(
y+y1
2
)+c=0
Hence, equation of polar of (1,1) is x.1+y.1+2(x+1)+3(y+1)−3=0 x+y+2x+2+3y+3−3=0 ⇒ 3x+4y+2=0