What is the standard form of the equation of a circle with a diameter that has endpoints (-8,0) and (4,-8)?

1 Answer
Feb 14, 2016

Answer:

#(x+2)^2 + (y+4)^2 = 52#

Explanation:

since the coords of the endpoints of the diameter are known, the centre of the circle can be calculated using the 'mid-point formula'.The centre being at the mid-point of the diameter.

centre = #[ 1/2(x_1+x_2) , 1/2(y_1+y_2)]#

let #(x_1 , y_1) = (-8 , 0)#
and#(x_2 , y_2) = (4 , -8 )#

hence centre # = [1/2(-8+4) ,1/2 (0-8 ) ] = ( -2 , -4 ) #

and radius is the distance from the centre to one of the end points. To calculate r , use the 'distance formula'.

# d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) #

let#(x_1 , y_1 ) = (-2 , -4 )#
and#(x_2 , y_2 ) = (-8 , 0 )#

hence r #= sqrt((-8+2)^2 + (0+4)^2) = sqrt(36+16) = sqrt52 #

centre = (-2 , -4 ) and # r = sqrt52#

the standard form of the equation of a circle is

# (x-a)^2 + (y-b)^2 = r^2 #
where (a , b ) are the coords of centre and r , is radius.

#rArr (x+2)^2 + (y+4)^2 = 52 #