How do you write the equation of the circle with a diameter that has endpoints at (8, 7) and (–4, –3)?

1 Answer
Jun 2, 2016

Answer:

Find the center and radius of the circle, and then write it in standard form as

#(x-2)^2+(y-2)^2=61#

Explanation:

The standard equation of a circle centered at #(x_0,y_0)# with radius #r# is #(x-x_0)^2+(y-y_0)^2=r^2#. Thus, if we can find the center and the radius, we can write the equation.

As the midpoint of any diameter of a circle is the center of the circle, we can find the center by locating the midpoint of the line segment with endpoints #(8,7)# and #(-4,-3)#. The midpoint of between any two points #(a_1,b_1)# and #(a_2,b_2)# is #((a_1+a_2)/2,(b_1+b_2)/2)#. Then, substituting our values, we have the midpoint, and thus the circle's center, at

#((8-4)/2,(7-3)/2)=(2,2)#

Next, to find the radius of the circle, we can just calculate the distance from the center to either of the given points on the circle. The distance between two points #(a_1,b_1),(a_2,b_2)# is #sqrt((a_2-a_1)^2+(b_2-b_1)^2)#. Then, using the center at the point #(8,7)#, we have the radius as

#sqrt((8-2)^2+(7-2)^2)=sqrt(36+25)=sqrt(61)#

With both the center at the radius, we can now write the circle's equation:

#(x-2)^2+(y-2)^2=(sqrt(61))^2#

#:. (x-2)^2+(y-2)^2=61#