TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Let S1:x2+y2−4=0
S2:(x−h)2+(y−k)2=‌

25

2

So, equation of common chord will be
S1−S2=0
⇒‌‌2hx+2ky−h2−k2=−‌

17

2

⇒2hx+2ky−h2−k2+‌

17

2

=0
Now, slope of common chord =‌

1

4

∴‌‌‌

−2h

2k

=‌

1

4

⇒k=−4h....(i)
Now, for maximum length of chord C1M=0
So, that AB become diameter of S1
|‌

−h2−k2+ 17 2

√4h2+4k2

|=0
⇒‌‌h2+k2=‌

17

2

.....(ii)
On solving Eqs. (i) and (ii), we get
(h,k)=(‌

√2

2

,∓2√2)