Question 984216
Not  a circle:  x2 +y2-6x+8y-24=0


This, a circle:  x^2+y^2-6x+8y-24=0


{{{x^2+y^2-6x+8y-24=0}}}
Complete the Square to put into standard form:  {{{(x-3)^2+(y+4)^2=49}}}.


The diameter must include the center of the circle, (3,-4), and the given point  (-1,1) must be on the LINE that contains a diameter of the circle  (as asked).


Same now as, find the equation for the line which passes through  (3,-4) and (-1,1).  Using point-slope form and only determining the slope of those two points, and choosing (-1,1),  equation is  {{{highlight(y-1=-(5/4)(x+1))}}}.