Asked by Anonymous
Circle C lies on the coordinate plane with its center at (2,−16).
Point A(2,−13) lies on circle C.
Which equation verifies that the point P(−1,−16) lies on circle C?
a. (2−16)^2+(2−13)^2=(2−16)^2+(−1−16)^2
b. (2+1)^2+(−16+16)^2=(2−2)^2+(−16+13)^2
c. (2+16)^2+(−1+16)^2=(2+2)^2+(−16−13)^2
d. (2−1)^2+(−16−16)^2=(2+2)^2+(−16−13)^2
Point A(2,−13) lies on circle C.
Which equation verifies that the point P(−1,−16) lies on circle C?
a. (2−16)^2+(2−13)^2=(2−16)^2+(−1−16)^2
b. (2+1)^2+(−16+16)^2=(2−2)^2+(−16+13)^2
c. (2+16)^2+(−1+16)^2=(2+2)^2+(−16−13)^2
d. (2−1)^2+(−16−16)^2=(2+2)^2+(−16−13)^2
Answers
Answered by
Reiny
circle with centre (2,-16) is
(x-2)^2 + (y+16)^2 = r^2
but A(2,−13) lies on it, so
0^2 + 3^2 = r^2 = 9
equation : (x - 2)^2 + (y + 16)^2 = 9
so to see if P(−1,−16) lies on it, I would try to show LS = RS
LS = (-1 - 2)^2 + (-16+16)^2
= 9 + 0 = 9
= RS , then yes, point P lies on the circle
Don't understand why they would want to make it look so complicated
(x-2)^2 + (y+16)^2 = r^2
but A(2,−13) lies on it, so
0^2 + 3^2 = r^2 = 9
equation : (x - 2)^2 + (y + 16)^2 = 9
so to see if P(−1,−16) lies on it, I would try to show LS = RS
LS = (-1 - 2)^2 + (-16+16)^2
= 9 + 0 = 9
= RS , then yes, point P lies on the circle
Don't understand why they would want to make it look so complicated
Answered by
alex
it’s b
Answered by
mia
whats the answer.
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