(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)

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.