How do you write an equation for a circle with diameter are P(−2, 2) and Q(6, −6)?

1 Answer
Mar 14, 2018

Answer:

I guess you meant end points of the diameter are #P(-2,2)# and #Q(6,-6)#.

Explanation:

First, you take out the midpoint of the line joining points # P# and #Q#.
Let it be #O#.
#O=((-2+6)/2,(2-6)/2)#
( Using midpoint formula)
#O=(4/2,-4/2)#
#O=(2,-2)#
Since, the midpoint of a diameter is the centre of the circle, and standard coordinate of the centre is #O(h,k)#,
We get #h=2# and #k=-2#.

Then, take out the length of the diameter using distance formula,
#sqrt((-2-6)^2+(2-(-6))^2)#
#sqrt(64+64)#
#sqrt(128)#
#sqrt(2^7)#
#2*2*2sqrt(2)#
#8sqrt2#
Now, #radius=(diameter)/2#
radius#=4sqrt2#

Now, we have #h=2#, #k=-2# and #r=4sqrt2#
Let us substitute these in the standard equation of a circle,
#(h-x)^2+(k-y)^2=r^2#
#(2-x)^2+(-2-y)^2=(4sqrt2)^2#
#4-4x+x^2+4+4y+y^2=32#
#x^2+y^2-4x+4y=24#
This is our required equation.

Hope this helps :)
Pretty long one, though.