Question

Show that (4,-2) and (3,-6) are conjugate with respect
to the circle x^2+y^2-24=0​

Answer 1

Step-by-step explanation:

$given \: \: {x}^{2} + {y}^{2} - 24 = 0 \\ so \\ a = {4}^{2} + {( - 2)}^{2} - 24 = - 4 \\ b = {3}^{2} + {( - 6)}^{2} - 24 = 21$

1st point lies inside the circle

and 2nd point lies outside the circle....so, they are conjugate point w.r.t. the circle.