(x-6)² + (x + 5)² = 16
in the xy-plane, the graph of the equation above is a
circle. point p is on the circle and has coordinates
(). if pq is a diameter of the circle, what are
the coordinates of point q?
a) (2, 5)
b) (6,-1)
c) (6,-5)
d) (6,-9)

answer A   : (2 -5)

Step-by-step explanation:

the midipoint P,Q is the center A of this circle (because PQ is a diameter)

use : (x-6)²+(x+5)² = 16 you have : A(6 , -5) now  let Q(x,y) :

 (x+10)/2 = 6

    (y-5)/2 = - 5                          

x+10 = 12

y-5 = -10

so : x = 2 and y = - 5answer A   : (2 -5)

right

step-by-step explanation: