The all conjugate of lines 5x+7y−78=0 with respect to the circle x2+y2+6x+8y−96=0 passes through the pole of the given line with respect to given circle. Let required pole be P(x1,y1) with respect to the circle T=0 xx1+yy1+3(x+x1)+4(y+y1)−96‌‌=0 (x1+3)x+(y1+4)y+(3x1+4y1−96)‌‌=0 The above line represent the line 5x+7y−78=0 only. Hence
x1+3
5
=
y1+4
7
=
3x1+4y1−96
−78
=k Therefore, x1=5k−3 y1=7k−4 And, 3x1+4y1=96−78k Thus, 3(5k−3)+4(7k−4)‌‌=96−78k 121k‌‌=121 k‌‌=1 So x1=2 y1=3 Therefore, the required point of concurrence of all conjugate lines of the given line with respect to the given circle is (2,3)