# How do you write the equation of the circle with a diameter that has endpoints at (7, –4) and (1, –10)?

##### 1 Answer

#### Answer:

#### Explanation:

The standard form of the equation of a circle is.

#color(red)(|bar(ul(color(white)(a/a)color(black)( (x - a)^2 + (y - b)^2 = r^2)color(white)(a/a)|)))# where (a ,b ) are the coordinates of the centre and r , the radius.

To obtain the equation , we require to find it's centre and radius.

Given the endpoints of the diameter , the centre will be at the mid-point and the radius will be the distance from the centre to either one of the endpoints.

Given 2 points

#A (x_1,y_1)" and "B (x_2,y_2) " the mid-point is "#

#color(blue)(|bar(ul(color(white)(a/a)color(black)( M_(AB) = 1/2(x_1 + x_2) , 1/2(y_1 + y_2))|)))# Here the 2 endpoints are (7,-4) and (1,-10)

hence centre =

# [ 1/2(7+1) , 1/2(-4-10) ] = (4 , -7) # To calculate the radius: distance from (4,-7) to (7,-4) using the

#color(blue)" distance formula " #

#color(green)(|bar(ul(color(white)(a/a)color(black)( d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2))color(white)(a/a)|)))#

where# (x_1,y_1)" and " (x_2,y_2)" are 2 coordinate points " # let

# (x_1,y_1)=(7,-4)" and " (x_2,y_2) =(4,-7)#

#rArr " radius " = sqrt((4-7)^2 + (-7+4)^2) = sqrt(9+9) #

# = sqrt18# We now have centre = (4 , -7) and radius =

#sqrt18#

hence equation of circle:

# (x - 4)^2 + (y + 7)^2 = 18 #