TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

We have, ∠APB=‌

π

4

⇒‌‌∠AOB=2∠APB=2×‌

π

4

=‌

π

2

∴‌‌∠OAB=∠OBA=‌

π

4

∴‌‌OA=OB⇒OA2=OB2
⇒‌‌(h+1)2=(h−5)2
⇒‌‌h2+2h+1=h2−10h+25
⇒‌‌12h=24⇒h=2
Since, ∠AOB=‌

π

2

∴‌‌OA⟂OB
⇒ Slope of OA× Slope of OB=−1
⇒‌‌‌

k−3

h+1

×‌

k−3

h−5

=−1
⇒(k−3)2=−(h+1)(h−5)
⇒k=0,6
∴ Centre of circle is (2,0) or (2,6)
When centre is (2,0)
∴ Radius of circle =OA
When centre is (2,6)
∴ Radius of circle =√(2+1)2+(6−3)2
=√9+9=3√2
∴ Equation of circle is
(x−2)2+(y−0)2=18
⇒‌‌x2+y2−4x−22=0
or (x−2)2+(y−6)2=18
⇒‌‌x2+y2−4x−12y+22=0