Given equation of circle:-
x2+y2+4x+6y+9=0
Given equation of line:-
3x+5y+17=0.....(1)
Let P(a,b) be the pole of line 3xx+5y+17=0 w.r.t. the circle x2+y2+4x+6y+9=0
Now equation of polar of point P(a,b) w.r.t. the circle x2+y2+4x+6y+9=0-
ax+by+2(x+a)+3(y+b)+9=0
⇒(a+2)x+(b+3)y+(2a+3b+9)=0.....(2)
As the equation (2)&(3) are same, therefore,
a+23=b+35=2a+3b+917
Now,
Case I:-
a+23=b+35
⇒5(a+2)=3(b+3)
⇒5a+10=3b+9
⇒3b−5a=1.....(3)
Case II:-
a+23=2a+3b+917
17(a+2)=3(2a+3b+9)
17a+34=6a+9b+27
9b−11a=7.....(4)
Multiplying eqn(3) by 3, we get
9b−15a=3.....(5)
Subtracting eqn(5) from (4), we have
(9b−11a)−(9b−15a)=7−3
9b−11a−9b+15a=4
4a=4
⇒a=1
Substituting the value of a in eqn(3), we have
3b−5(1)=1
⇒3b=1+5
⇒b=2
Hence the the pole of the line 3x+5y+17=0 w.r.t. the circle x2+y2+4x+6y+9=0 is (1,2).