How do you write the equation in standard form of the circle for; endpoints of the diameter A(14, -7) and B(-10,9)?

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Answer 1 · 2016-01-12T10:13:53

#(x-2)^2+(y-1)^2=208#

The general form of a circle centred at #(a,b)# and with radius #r# is #(x-a)^2+(y-b)^2=r^2#.

So we first need to find the radius and the centre before we can find the equation.

Length of diameter will be distance between its endpoints.

#therefore d=sqrt((14-(-10))^2+(-7-9)^2)=sqrt832#.

Therefore the radius #r=1/2d=1/2sqrt832#.

This implies that #r^2=832/4=208#.

The centre of the circle will be the midpoint of the diameter, ie

#(a,b)=((14-10)/2;(9-7)/2)=(2,1)#.

Therefore the standard equation of this circle will be

#(x-2)^2+(y-1)^2=208#.