TS EAMCET 19-July-2022 Shift 2 Solved Paper

Given, equation of circle is
x2+y2−10x+14y+46=0 and equation of line is 3x−5y+6=0
Let P(α,β) be the pole of line
3x−5y+6=0......(i)
Equation of polar of point P w.r.t. the circle, we get
x2+y2−10x+14y+46=0 is
Comparing Eqs. (i) and (ii),
‌

α−5

3

=‌

β+7

−5

=‌

−5α+7β+46

6

⇒−5α+25=3β+21⇒−5α=3β−4
∴‌

β+7

−5

=‌

3β−4+7β+46

6

⇒6β+42=−50β+42×(−5)
⇒−5α=−‌

35

2

⇒α=‌

7

2

∴‌α+β=‌

7

2

−‌

9

2

=−‌

2

2

=−1