Question

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

Answer 1

(5−6)2+(−15+12)2=(5−2)2+(−15+14)2

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