If the radical axis of the circles `x^2+y^2+2gx+2fy+c=0` and `2x^2+2y^2+3x+8y+2c=0` touches the circle `x^2+y^2+2x-2y+1=0` , show that either `g=3/4` or `f=2`

Options

A. `g=(4)/(3)` and f=2
B. `g=(4)/(3)` and `f=(1)/(2)`
C. `g=-(3)/(4)` and `f=2`
D. `g=(3)/(4)` and `f=(1)/(2)`

Answer

B

Explanation

Equations of the two circles are
`x^(2)+y^(2)+2gx+2fy+c=0 and x^(2)+y^(2)+(3)/(2)x+y+c=0`
The equation of radical axis is
`(2g-(3)/(2))x+(2f-1)y=0` ...(i)
It touches the circle `x^(2)+y^(2)+2x+2y+1=0`
`:. |(-(2g-(3)/(2))-(2f-1))/(sqrt((2g-(3)/(2))^(2)+(2f-1)^(2)))|=1`
`rArr(2g+2f-(5)/(2))^(2)=(2g-(3)/(2))^(2)+(2f-1)^(2)`
Clearly `g=(4)/(3) and f=(1)/(2)` satisfy this relation.