TS EAMCET 18-July-2022 Shift 1 Solved Paper

Let S1:x2+y2+4y=0,C1(0,−2),r1=2 and S2:x2+y2−4x−5=0,C2(2,0) ,
r2=3

Common chord of S1 and S2 is
S1−S2=0
4x+4y+5=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
M is the point of intersection of common chord and line joining C1 and C2 .
Let M divides C1C2 in λ:1 ratio internally.
M(‌

2λ

λ+1

,‌

−2

λ+1

)
M lies on common chord
‌

8λ

λ+1

−‌

8

λ+1

+5‌=0⇒8λ−8+5λ+5=0
λ‌=‌

3

13

Also M is the centre of the circle whose diameter is common chord of S1 and S2
So, abscissa of M=‌

2λ

λ+1

=‌

‌ 6 13

3 13 +1

=‌

6

16

=‌

3

8