If the lines 2 x+y+12=0, k x-3 y-10=0 are conjugate with respect to the circle x^2+y^2-4 x+3 y-1=0 , then k=

Step-by-step solution

Lines 2 x+y+12=0, k x-3 y-10=0 are conjugate with respect to given circle. Pole of line 2 x+y+12=0 lie on the line k x-3 y-10=0 Equation of polar of pole (x, y) to circle x^2+y^2-4 x+3 y-1=0 is array r x x₁+y y₁-2 (x+x₁ )+ 3 2 (y+y₁ )-1=0 x (x₁-2 )+y (y₁+ 3 2 )-2 x₁+ 3 2 y₁-1=0 array it is same as 2 x+y+12=0 So, x₁-2 2 = y₁+ 3 2 1 = -2 x₁+ 3 2 y₁-1 12 x₁-2=2 y₁+3 and 12 y₁+18=-2 x₁+ 3 2 y₁-1 aligned ( 21 2 +4 ) y₁ & =-29 ( 29 2 ) y₁ & =-29 y₁=-2 aligned put in Eq. (i) x₁+4=5 x₁=1 Point (x₁, y₁ ) lie of k x-3 y-10=0