Question 1084215: In the (x,y) coordinate plane, which of the following is an equation of the circle having the points (3,-3) and (-5,5) as endpoints of a diamteter?
F) (x+1)^2 + (y-1)^2 = sqrt 32 G) (x+1)^2 + (y+1)^2 = sqrt 32 H) (x-1)^2 + (y+1)^2 = sqrt 32
J) (x-1)^2 + (y+1)^2 = 32 K) (x+1)^2 + (y-1)^2 = 32
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Need to know the center and the radius.
Midpoint of diameter is center, which is midpoint formula, which gives (-1, 1) and the radius is the distance from the center to either end of the diameter. That is the distance formula, sqrt ((x2-x1)^2+(y2-y1)^2)= sqrt (16+16)=sqrt (32)
so r^2=32. The equation of a circle is x^2+y^2=r^2. So once one has the sqrt of the radius (here sqrt (32)), it has to be squared (32) to be part of the equation.
(x+1)^2+(y-1)^2=32
K.
|
|
|
