From the point of intersection of the circle S: x^2+y^2-4 x+6 y+13=0 and the line L: 2 x+5 y+11=0 two tangents are drawn to the circle x ^2+ y ^2= 121 29 , whose slopes are m ₁ and m ₂ , then -

Step-by-step solution

S:(x-2)^2+(y+3)^2=0 S is a point circle which represents point (2,-3) and this point (2,-3) also lies on the line L: 2 x+5 y+11=0 Equation of tangents from (2,-3) to the circle aligned & x ^2+ y ^2= 121 29 is y +3= m ( x -2) & mx - y =3+2 ~m aligned Applying p = r aligned & | 3+2 ~m 1+ m ^2 |= 121 29 & 29 (9+12 ~m +4 ~m ^2 )=121 (1+ m ^2 ) & 5 ~m ^2-348 ~m -140=0 & ~m ₁+ m ₂= 348 5 aligned & m₁ m₂=- 140 5 =-28