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

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