The minimum radius of the circle which contains the three circles, `x^(2)+y^(2)-4y-5=0,x^(2)+y^(2)+12x+4y+31=0` and `x^(2)+y^(2)+6x+12y+36=0` is

Options

A. `(7)/(18)sqrt(900)+3`
B. `(sqrt(845))/(9)+4`
C. `(5)/(36)sqrt(949)+3`
D. none of these

Answer

3

Explanation

`S_(1)-=x^(2)+y^(2)-4y-5=0`
`S_(2)-=x^(2)+y^(2)+12x+4y+31=0`
and `S_(3) -=x^(2)+y^(2)+6x+12y+36=0`
`C_(1) -= (0,2),r_(1)=3`
`C_(2) -= (-6,-2), r_(2)=3`
`C_(3) -= (-3,-6) , r_(3)=3`

All circles have the same radius,
The radius of the circle touching all the circles is
`CP=C C_(1)=C_(1)P+3`
Let `C(h,k)` be the center of the required circle. Then,
`C C_(1)= C C_(2) = C C_(3)` or `C C_(1)^(2) = C C_(2)^(2)= C C_(3)^(2)`
or `(h-0)^(2)+(k-2)^(2)=(h+6)^(2)+(k+2)^(2)`
`=(h+3)^(2)+(k+6)^(2)`
or `-4k+4=12h+4k+40=6h+12k+45`
or `3h+2k+9=0` and `6h-8k-5=0`
Solving, we get
`h=(-31)/(18),k=(-23)/(12)`
`:. C C_(1)=sqrt((0+(31)/(18))^(2)+(2+(23)/(12))^(2))=(5)/(36)sqrt(949)`
Now, `CP =C C_(1)+3=(5)/(36)sqrt(949)+3`