(a) the midpoint is just the average of the two end points: (6,4)
(b) the radius is half the length of the line segment. Use your normal distance formula.
(c) you have the center and radius. You can see that x=24 is too far from the center.
A scientist is plotting the circular path of a partial on a coordinate plane for a lab experiment. The scientist knows the path is a perfect circle and that the particles start at ordered pair (-5, -11). When a particle is halfway down the circle, the particle is the ordered pair (11, 19). The segment formed by connecting these points has a center of the circle midpoint. a) What is the ordered pair that represents the center of the circle? b) What is the length of the radius, in units, of the circle? c) Explain why the the particle can never pass through a point with an x-coordinate of 24 as long as it stays on a circular path.
3 answers
" start at ordered pair (-5, -11). When a particle is halfway down the circle, the particle is the ordered pair (11, 19). "
This tells me that the two points in the above sentence form the diameter
of the circle, so the center would be midpoint, which is
( (-5+11)/2 , (-11+19)/2 )
= (3,4)
so the circle is (x-3)^2 + (y-4)^2 = r^2
but (-5,-11) lies on this, so
(-5-3)^2 + (-11-4)^2 = r^2
289 = r^2
r = 17
circle is (x-3)^2 + (y-4)^2 = 289
if x = 24
441 + (y-4)^2 = 289
(y-4)^2 = -152
so y-4 is not a real number, thus y is not a real number
This tells me that the two points in the above sentence form the diameter
of the circle, so the center would be midpoint, which is
( (-5+11)/2 , (-11+19)/2 )
= (3,4)
so the circle is (x-3)^2 + (y-4)^2 = r^2
but (-5,-11) lies on this, so
(-5-3)^2 + (-11-4)^2 = r^2
289 = r^2
r = 17
circle is (x-3)^2 + (y-4)^2 = 289
if x = 24
441 + (y-4)^2 = 289
(y-4)^2 = -152
so y-4 is not a real number, thus y is not a real number
dang! forgot to divide 6 by 2.
Go with mathhelper's usual careful work.
Go with mathhelper's usual careful work.