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

##### 1 Answer

#### 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

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 #