KCET2012MathematicsCircle
The equations of the two tangents from (-5,-4) to the circle x²+y²+4 x+6 y+8=0 are
Options
- Ax+2 y+13=0,2 x-y+6=0
- B2 x+y+13=0, x-2 y=6
- C3 x+2 y+23=0,2 x-3 y+4=0
- Dx-7 y=23,6 x+13 y=4
Correct answer
A. x+2 y+13=0,2 x-y+6=0
Step-by-step solution
Any line through the point (-5,-4) is aligned y+4 &=m(x+5) m x-y &+(5 m-4)=0 aligned Now, radius of circle = (2)²+(3)²-8 = 4+9-8 = 5 If it is a tangent, then perpendicular from centre (-2,-3) is equal to the above radius. m(-2)-(-3)+(5 m-4) m²+1 = 5 -2 m+3+5 m-4= 5 1+m² 3 m-1= 5 1+m² (3 m-1)²=5 (1+m² ) 9 m²+1-6 m=5+5 m² 4 m²-6 m-4=0 4 m²-8 m+2 m-4=0 4 m(m-2)+2(m-2)=0 (m-2)(4 m+2)=0 m =2,- 1 2 Putting the value of m=2 in Eq. (i), we get aligned 2 x-y+5 2-4 &=0 2 x-y+6 &=0 aligned Again, putting the value of m =- 1 2