A circle has a diameter with end points (3,-2) and (9,-8) what is the equation of the circle?
My options:
R^2=(x+6)^2+(y-5)^2
R^2=(x+6)^2+(y-3)^2
R^2=(x-6)^2+(y+3)^2
R^2=(x-6)^2+(y+5)^2

Question

A circle has a diameter with end points (3,-2) and (9,-8) what is the equation of the circle? 
My options: 
R^2=(x+6)^2+(y-5)^2
R^2=(x+6)^2+(y-3)^2
R^2=(x-6)^2+(y+3)^2
R^2=(x-6)^2+(y+5)^2

mertsj · Answer

Find the midpoint of the diameter.  That will be the center of the circle.

anonymous · Answer

Equation of any circle centred on the point (a,b) with a radius r is given by:$$(x-a)^2 + (y-b)^2 = r^2$$ So in your case, the radius is just R (still unknown). And we are left to find the centre ourselves. |dw:1364255207325:dw| We see that the centre must be the middle of this line one the diagram. So taking the average of the x and y values gives us (9+3)/2=6, and (-2-8)/2=-5. So the centre is (6,-5). From here plug it back into the formula and we get the formula to be $$(x-6)^2 + (y+5)^2 = R^2$$

anonymous · Answer

Thank you!