AP EAPCET 08-Jul-2022 Shift 2 Paper

S1≡x2+y2=4...(i)
S2≡x2+y2−2x−3=0...(ii)
S3≡x2+y2−2y−3=0...(iii)
Let S4≡x2+y2+2gx+2fy+c be the circle which is orthogonal with the circles S1,S2 and S3
Condition of orthogonal,
2(gg′+ff′)=c+c′
Now, for S1 and S4 ,
g×0+f×0=˙c−4
⇒c=4
For S2 and S4
2[g×(−1)+f×(0)]=4−3
⇒g=−

1

2

For S3 and S4
2[g×(0)+f×(−1)]=4−3
⇒f=−

1

2

Now, radius of S4
r2=g2+f2−c=(

1

4

+

1

4

−4)=−

7

2

<0
Thus, no such S4 circle is possible which is orthogonal to S1,S2,S3