TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given S1=x2+y2=16
Centre (0,0)r=4Clearly, the diameter of S1 will be common chord.
Let PQ be common chord and centre of S2 be (h,k)
We have AP=5,PB=4
∴‌‌AB=3
Where B=(0,0)
Slope of PQ=‌

3

4

∴ Slope of AB=‌

−4

3

By parameter equation
‌

h−0

−3 5

‌‌=‌

k−0

4 5

=±3
x‌‌=∓‌

9

5

,y=‌

±12

5

∴‌ Centre ‌(‌

−9

5

,‌

12

5

)‌ and ‌(‌

9

5

,‌

−12

5

)