How do you write an equation for a circle with endpoints of a diameter at the points (7,4) and (-9,6)?

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Answer 1 · 2016-10-25T18:30:12

$(x+1)^2+(y-5)^"=65$

The standard equation of a circle is: $(x-a)^2+(y-b)^"=r^2$
where $(a, b)$ are the co-ordinates of the centre and $r$ is the radius.

Endpoints of diameter $(7, 4)$ & $(-9, 6)$

The centre is the MID-POINT of a diameter

$(a, b)= ((7+(-9))/2, (4+6)/2)$

giving $(a, b)=(-1, 5)$

To find the radius use Pythagoras on one end point of diameter and the centre.

$r=sqrt((-1-7)^2+(5-4)^2)$

$r=sqrt65$

so circle equation is;

$(x+1)^2+(y-5)^"=65$