Let each of the circles
`S_(1)-=x^(2)+y^(2)+4y-1=0`
`S_(1)-= x^(2)+y^(2)+6x+y+8=0`
`S_(3)-=x^(2)+y^(2)-4x-4y-37=0`
touch the other two. Also, let `P_(1),P_(2)` and `P_(3)` be the points of contact of `S_(1)` and `S_(2) , S_(2)` and `S_(3)`, and `S_(3)` , respectively, `C_(1),C_(2)` and `C_(3)` are the centres of `S_(1),S_(2)` and `S_(3)` respectively.
`P_(2)` and `P_(3)` are images of each other with respect to the line

Options

A. `y=x`
B. `y = -x`
C. `y=x+1`
D. `y= -x+2`

Answer

1

Explanation

`S_(1)-= x^(2)+y^(2)+4y-1=0`
`S_(2) -= x^(2)+y^(2)+6x+y+8=0`
`S_(3)-=x^(2)+y^(2)-4x-4y-37=0`
`C_(1) -= (0,-2),r_(1)=sqrt(5)`
`C_(2) -= (-3,(-1)/(2)),r_(2)=(sqrt(5))/(2)`
`C_(3)-= (2,2), r_(3)= 3 sqrt(5)`
Also, `C_(1)C_(2)=sqrt(9+(9)/(4))=(3sqrt(5))/(2)=r_(1)+r_(2)`
So, `S_(1)` and `S_(2)` touch each other externally,
`C_(2)C_(3)=sqrt(25+(25)/(4))=(5sqrt(5))/(2)=r_(3)-r_(2)`
So, `S_(2)` and `S_(3)` touch each other internally.
The point of contact `P_(1)` divides `C_(1)C_(2)` internally in the ratio `r_(1) : r_(2) = 2:1`
`implies P_(1) -= (-2 ,-1)`
The point of contact `P_(2)` divides `C_(2)C_(3)` externally in the ratio `r_(2) : r_(3) = 1:6`
`implies P_(2) -= (-4, -1)`.
The point of contact `P_(3)` divides `C_(3)C_(1)` externally in the ratio `r_(3) : r_(1) = 3:1`
`implies P_(3) -= ( -1,-4)`
Area of `Delta P_(1)P_(2)P_(3)= (1)/(2) | {:(-2,-1,1),(-4,-1,1),(-1,-4,1):}| = 3`
And area `Delta C_(1)C_(2)C_(3)= (1)/(2) |{:(0,-2,1),(-3,(-1)/(2),1),(2,2,1):}|=(15)/(2)`
`:. ("area "(Delta P_(1)P_(2)P_(3)))/("area"(DeltaC_(1)C_(2)C_(3)))=(3)/(15)=2:5`
Clearly, `P_(2)(-4,-1)` and `P_(3)(-1,-4)` are images of each other with respect to the line `y=x`