AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The pole of the line is taken as (h,k) of the line 9x+y−28=0 with respect to the circle.
x2+y2−

3

2

x+

5

2

y−

7

2

=0
The equation of polar becomes.
hx+ky−

3

4

(x+h)+

5

4

(y+k) −

7

2

=0
x(h−

3

4

)+y(k+

5

4

)−

3

4

h+

5

4

k −

7

2

=0
x(4h−3)+y(4k+5)−3h +5k−14=0
The equation shows the equation of line.

4h−3

9

=

4k+5

1

=

−3h+5k−14

−28

=λ
h=

3+9λ

4

,k=

λ−5

4

, 3h+5k−14=−28λ
On simplifying the above we get,
−3(

3+9λ

4

)+5(

λ−5

4

)−14=−28λ
−22λ−90=−11λ
90λ=90
λ=1
Thus, the pole of the given line is (3,-1)