TS EAMCET 6-Aug-2021 Shift 2 Question Paper

The equation of polar w.r.t the point (1,1) to the circle x2+y2−4x−6y+12=0 is
x⋅1+y⋅1−2(x+1)−3(y+1)+12=0
⇒‌‌−x−2y+7=0
⇒‌‌x+2y−7=0
Since, the inverse of the point (1,1) is the foot (h,k) of the perpendicular from the point (1,1) to the line x+2y−7=0
So, ‌‌‌

h−1

1

=‌

k−1

2

=‌

−(1+2−7)

(1)2+(2)2

⇒‌‌h−1=‌

4

5

⇒h=‌

9

5

and ‌‌‌

k−1

2

=‌

4

5

⇒k−1=‌

8

5

⇒‌‌k=13∕5
∴‌‌h+k=‌

9

5

+‌

13

5

=‌

22

5