Question

Pole of 3x+ 5y + 17 =0 with respect to the circle x^2+ y^2 + 4x + 6y +9 = 0 is​

Answer 1

Step-by-step explanation:

Given circle is x

2

+y

2

+4x+6y+9=0 and given line is 3x + 5y + 17 = 0

Let P(α,β) be the pole of line (ii) with respect to circle (i)

Now equation of polar of point P(α,β) with respect to circle (i) is

xα+yβ+2(x+α)+3(y+β)+9=0

or (α+2)x+(β+3)y+2α+3β+9=0

Now lines (ii) and (iii) are same therefore

3

α+2

=

5

β+3

=

17

2α+3β+9

From (i) and (ii) we get 5α+10=3β+9or5α−3β=−1.......(iv)

From (i) and (iii) we get 17α+34=6α+9β+27or11α−9β=−7........(v)

Solving (iv) & (v) we get α=1,β=2 Hence required pole is (1, 2)

Answer 2

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Find the pole of the line 3x+5y+17=0 with respect to the circle x

2

+y

2

+4x+6y+9=0

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ANSWER

Given circle is x

2

+y

2

+4x+6y+9=0 and given line is 3x + 5y + 17 = 0

Let P(α,β) be the pole of line (ii) with respect to circle (i)

Now equation of polar of point P(α,β) with respect to circle (i) is

xα+yβ+2(x+α)+3(y+β)+9=0

or (α+2)x+(β+3)y+2α+3β+9=0

Now lines (ii) and (iii) are same therefore

3

α+2

=

5

β+3

=

17

2α+3β+9

From (i) and (ii) we get 5α+10=3β+9or5α−3β=−1.......(iv)

From (i) and (iii) we get 17α+34=6α+9β+27or11α−9β=−7........(v)

Solving (iv) & (v) we get α=1,β=2 Hence required pole is (1, 2)