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#