benhimmel88
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Circle O is centered at (5,−15) and point A(2,−14) lies on ⨀O.
Which answer verifies that the point P(6,−12) is on the circle?

(5−6)2+(−15+12)2=(5−2)2+(−15+14)2

(5−6)2+(15+12)2=(5−2)2+(15+14)2

(6+12)2+(5+15)2=(5+15)2+(2+14)2

(6−12)2+(5−15)2=(5−15)2+(2−14)2

Respuesta :

mysticchacha

Answer:

(5−6)2+(−15+12)2=(5−2)2+(−15+14)2

Step-by-step explanation:

Circle equation is :  (x -h)^2 + (y - k)^2 = r^2

center (h, k) = (5, -15)

r = Distance from (5, -15) to point A (2, -14)

r = root ( (5 - 2)^2 + (-15 - (-14))^2 )

r = root ( 3^2  + 1^2)

r = root(10)

Equation is   (x - 5)^2 + (y + 15)^2 = 10

Check point P(6, -12)

(6  - 5)^2  + ( (-12) + 15 )^2  check if this equals 10

1^2 +  3^2 = 1 + 9 = 10,   10 = 10  good

It is the same as using the equation:

(5−6)2+(−15+12)2=(5−2)2+(−15+14)2    

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