The value of k so that `x^(2)+y^(2)+kx+4y+2=0` and `2(x^(2)+y^(2))-4x-3y+k=0` cut orthogonally, is

Options

A. `(10)/(3)`
B. `(-8)/(3)`
C. `(-10)/(3)`
D. `(8)/(3)`

Answer

C

Explanation

The equations of the circles are
`x^(2)+y^(2)+kx+4y+2=0` ...(i)
`x^(2)+y^(2)-2x-(3)/(2)y+(k)/(2)=0` ...(ii)
These two circles will interset orthogonally, if
`2((k)/(2)xx(-1)+2xx(-(3)/(4)))=2+(k)/(2) [:' 2 (g_(1)g_(2) +f_(1)f_(2))=c_(1)+c_(2)]`
`rArr -k-3=2+(k)/(2)rArr (3)/(2)k=-5 rArr k = (-10)/(3)`