Question

Find the pole of the line x+y+2=0 with respect to the circle x^2+y^2-4x+6y-12=0

Answer 1

Step-by-step explanation:

Given equation of polar-

x+y−2=0.....(1)

Let P(h,k) be the pole of line x+y=2 w.r.t the circle x

2

+y

2

−4x+6y−12=0

Therefore the polar of P w.r.t. the circle is-

hx+ky−2(x+h)+3(y+k)−12=0

(h−2)x+(k+3)y−2h+3k−12=0.....(2)

Now equation (1)&(2) are same.

Therefore,

1

h−2

=

1

k+3

=

2

−2h+3k−12

⇒4h−3k+8=0.....(3)

⇒2h−k+18=0.....(4)

Multiplying equation (4) by 3, we get

6h−3k+54=0.....(5)

Subtracting equation (3) from (5), we have

(6h−3k+54)−(4h−3k+8)=0

⇒2h+46=0

⇒h=−23

Substituting the value of h in equation (4), we have

−46−k+18=0

k=−28

Hence the pole of line x+y+2=0 w.r.t. the circle x

2

+y

2

−4x+6y−12=0 is (−23,−28).