Asked by johannes
find the equation of the circle having the line segment joining the points (-2,2) and (-8,6)as diameter.
Answers
Answered by
Damon
center halfway between
average x = -10/2 = -5
average y =8/2= 4
so center at (-5 , 4)
form therefore is
(x+5)^2 + (y-4)^2 = r^2
put point in to find r
(-2+5)^2 + (2-4)^2 = r^2
9 + 4 =r^2
r ^2 = 13
so
(x+5)^2 + (y-4)^2 = 13
========================
check, other point, y = 6
(x+5)^2 + 4 = 13
(x+5)^2 = 9
X+5 = 3 or -3
-3 ok, x = -8 as advertized
average x = -10/2 = -5
average y =8/2= 4
so center at (-5 , 4)
form therefore is
(x+5)^2 + (y-4)^2 = r^2
put point in to find r
(-2+5)^2 + (2-4)^2 = r^2
9 + 4 =r^2
r ^2 = 13
so
(x+5)^2 + (y-4)^2 = 13
========================
check, other point, y = 6
(x+5)^2 + 4 = 13
(x+5)^2 = 9
X+5 = 3 or -3
-3 ok, x = -8 as advertized
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