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

Given: 2 x + y + 12 = 0       . . . i k x - 3 y - 10 = 0     . . . i i Since, above lines are conjugate w.r.t. to the circle then pole of one passes through another. Let pole is h , m . Now, x 2 + y 2 - 4 x + 3 y - 1 = 0 T = 0 ⇒ h x + m y - 4 x + h 2 + 3 y + m 2 - 1 = 0 ⇒ h - 2 x + m + 3 2 y + - 2 h + 3 m 2 - 1 = 0     . . . i i i And by comparing i   &   i i i , we get h - 2 2 = m + 3 2 1 = - 2 h + 3 m 2 - 1 12 By solving, we get h - 2 m = 5   &#160