ranahussein774
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(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
(10,--5). 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 :

ghanami

Answer:

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 = - 5....answer A   : (2 -5)

Answer:

it (2,-5)

Step-by-step explanation:

just did it

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