The endpoints of the diameter of a circle are (-4,-5) and (-2,-1). What is the center, radius, and equation?

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Answer 1 · 2016-08-27T18:03:57

The Centre is#(-3,-3), "radius r"=sqrt5#.

The eqn.# : x^2+y^2+6x+6y+13=0#

Let the given pts. be #A(-4,-5) and B(-2,-1)#

Since these are the extremities of a diameter, the mid-pt. #C# of segment #AB# is the centre of the circle.

Hence, the centre is #C=C((-4-2)/2, (-5-1)/2)=C(-3,-3)#.

#r "is the radius of the circle" rArr r^2=CB^2=(-3+2)^2+(-3+1)^2=5#.

#:. r=sqrt5#.

Finally, the eqn. of the circle, with centre #C(-3,-3)#, and radius#r#, is

#(x+3)^2+(y+3)^2=(sqrt5)^2, i.e., x^2+y^2+6x+6y+13=0#