The equation of the two tangents from (-5,-4) to the circle `x^(2)+y^(2)+4x+6y+8=0` are

Options

A. `x+2y+13=0,2x-y+6=0`
B. `2x+y+13=0,x-2y=6`
C. `3x+2y+23=0,2x-3y+4=0`
D. `x-7y=23,6x+13y=4`

Answer

A

Explanation

Any line through the point (-5,-4)
is
`y+4=m(x+5)`
`impliesmx-y+(5m-4)=0` . . . (i)
It it is tangent, then perpendicular from centre (-2,-3) is equal to radius.
`therefore`Radius=`sqrt((2)^(2)+(3)^(2)-8)`
`=sqrt(4+9-8)=sqrt(5)`
`implies(m(-2)-(-3)+(5m-4))/(sqrt(m^(2))+1)=sqrt(5)`
`implies-2m+3+5m-4=sqrt(5)sqrt(m^(2)+1)`
`implies(3m-1)^(2)=(5m^(2)+1)`
`implies9m^(2)+1-6m=5m^(2)+5`
`implies4m^(2)-6m-4=0`
`implies2m^(2)-3m-2=0impliesm=-(1)/(2),2`
`therefore` Reqruied equations are
`2x-y+6=0` ltbgrgt and x+2y+13=0.