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Question

Find the pole of the line 3x + 5y + 17 =0 with respect to the circle x2+y2+4x+6y+9=0

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Solution

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
3b5a=1.....(3)
Case II:-
a+23=2a+3b+917
17(a+2)=3(2a+3b+9)
17a+34=6a+9b+27
9b11a=7.....(4)
Multiplying eqn(3) by 3, we get
9b15a=3.....(5)
Subtracting eqn(5) from (4), we have
(9b11a)(9b15a)=73
9b11a9b+15a=4
4a=4
a=1
Substituting the value of a in eqn(3), we have
3b5(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).

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